Q: x-अक्ष के बिन्दुओं x=−a तथा x=a में प्रत्येक पर समान आवेश q रखा हैं, तथा इसके केन्द्र पर m द्रव्यमान तथा q0=2q आवेश का एक कण रखा हैं। यदि आवेश q0 को y-अक्ष के अनुदिश अल्प दूरी (y<<a) विस्थापित किया जाए, तो कण पर लगने वाला परिणामी बल समानुपाती होगा,

Fnet=(y2+a2)22kq(2q)⋅y2+a2yFnet=(y2+a2)3/22kq(2q)y⇒a3kq2yअतः F∝y

Q: कल्पना कीजिये कि एक प्रोटॉन और एक इलेक्ट्रान के आवेश में अल्प अन्तर होता है। इनमें से एक −e है और दूसरा (e+Δe) है। यदि एक दूसरे से ' d ' दूरी पर रखे हाइड्रोजन के दो परमाणुओं के बीच (जहाँ d परमाणु के साइज से बहुत अधिक है) स्थिर वैद्युत बल और गुरूत्वीय बल का परिणामी (नेट) शून्य है तो, Δe की कोटि होगी : (दिया है, हाइड्रोजन का द्रव्यमान mh=1.67×10−27 kg )

प्रश्नानुसार, स्थिर वैद्युत बल का परिणामी (FE)= गुरुत्वीय बल का परिणामी (FG)FE=FGया 4πε01 d2Δe2= d2Gm2⇒Δe=mKG(4πε01=k=9×109)=1.67×10−279×1096.67×10−11Δe≈1.436×10−37C

Q: किसी उभयनिष्ठ बिन्दु से, लम्बाई l की दो द्रव्यमानहीन डोरियों से निलंबित, दो सर्वसम आवेशित गोले, अन्योन्य प्रतिकर्षण के कारण, आरम्भ में एक-दूसरे से d(d<<1) दूरी पर हैं। दोनों ही गोलों से एक नियत दर से आवेशों का क्षरण आरम्भ होता है, और इसके परिणामस्वरूप गोले एक दूसरे की ओर वेग v से आते हैं। तब गोलों के बीच की दूरी, x के फलन के रूप में वेग v का विचरण किस रूप में होता है?

चित्र से, tanθ=mgFe≃θया x3∝q2.........(i) या x3/2∝q.........(ii)समय के सापेक्ष समीकरण (1) का अवकलन करने पर3x2dtdx∝2qdtdq परन्तु dtdq= स्थिरांकअत: x2(v)∝q समीकरण (2) से q का मान रखने पर x2(v)∝x3/2 या v∝x−1/2

Q: In the following diagram the work done in moving a point charge from point P to point A,B and C is respectively as WA′WB and WC′ then

According to the figure, there is no other charge. A stationary charge when moved in a space of no field, does not apply any force. No work is doneWA=WB=WC=0

Q: An object A has a charge of −2μC and the object B has a charge of +6μC. Which statement is true?

Vector from of Coulomb's law states that FAB and FBA are equal and opposite.Since, r^AB and r^BA are unit vectors pointing in opposite direction, we haver^AB=−r^BATherefore, equations of force is Given byFAB=4πε01r2q1q2(r^AB)=−4πε01r2q1q2r^BA=−FBA

Q: .Three charges each of magnitude q are placed at the corners of an equilateral triangle, the electrostatic force on the charge placed at the center is (each side of triangle is L )

In the following figure since ∣∣FA∣∣=∣∣FB∣∣=∣∣FC∣∣ and they are equally inclined with each other, so their resultant will be Zero

Q: When the distance between the charged particles is halved, the force between them becomes

F∝r21; so when r is halved the force becomes four times

Q: If a conducting medium is placed between two charges, then the electric force between them will become.

When a conducting medium is placed between the two charges, then both charges combine in such a way that their combination behaves like a single charge. Electrical force acts only between two different charges. But, due to the absence of two different charges after the combination of charges with conducting medium,the force between them becomes zero.

Q: A uniformly charged thin spherical shell of radius R carries uniform surface charge density of σ per unit area. It is made of two hemispherical shells, held together by pressing them with force F (see figure). F is proportional to

Electrostatic repulsive force;Fele =(2ε0σ2)πR2;F=Fele =2ε0σ2πR2

Q: Two point charges +9e and +e are at 16 cm away from each other. Where should another charge q be placed between them so that the system remains in equilibrium

For equilibrium statex29×qc=(16−x)2q×e or, x29=(16−x)21 or, x3=16−x1or, 3(16−x)=xor, 48−3x=xor, 4x=48 or, x=12 (Ans) Now, from x the distance will be (16−12)cm=4 cm.

Question 1

x-अक्ष के बिन्दुओं x=−a तथा x=a में प्रत्येक पर समान आवेश q रखा हैं, तथा इसके केन्द्र पर m द्रव्यमान तथा q0​=2q​ आवेश का एक कण रखा हैं। यदि आवेश q0​ को y-अक्ष के अनुदिश अल्प दूरी (y<<a) विस्थापित किया जाए, तो कण पर लगने वाला परिणामी बल समानुपाती होगा,

Accepted Answer

Fnet​=(y2+a2​)22kq(2q​)​⋅y2+a2​y​Fnet​=(y2+a2)3/22kq(2q​)y​⇒a3kq2y​अतः F∝y

Question 2

कल्पना कीजिये कि एक प्रोटॉन और एक इलेक्ट्रान के आवेश में अल्प अन्तर होता है। इनमें से एक −e है और दूसरा (e+Δe) है। यदि एक दूसरे से ' d ' दूरी पर रखे हाइड्रोजन के दो परमाणुओं के बीच (जहाँ d परमाणु के साइज से बहुत अधिक है) स्थिर वैद्युत बल और गुरूत्वीय बल का परिणामी (नेट) शून्य है तो, Δe की कोटि होगी : (दिया है, हाइड्रोजन का द्रव्यमान mh​=1.67×10−27 kg )

Accepted Answer

प्रश्नानुसार, स्थिर वैद्युत बल का परिणामी (FE​)= गुरुत्वीय बल का परिणामी (FG​)FE​=FG​या 4πε0​1​ d2Δe2​= d2Gm2​⇒Δe=mKG​​(4πε0​1​=k=9×109)=1.67×10−279×1096.67×10−11​​Δe≈1.436×10−37C

Question 3

किसी उभयनिष्ठ बिन्दु से, लम्बाई l की दो द्रव्यमानहीन डोरियों से निलंबित, दो सर्वसम आवेशित गोले, अन्योन्य प्रतिकर्षण के कारण, आरम्भ में एक-दूसरे से d(d<<1) दूरी पर हैं। दोनों ही गोलों से एक नियत दर से आवेशों का क्षरण आरम्भ होता है, और इसके परिणामस्वरूप गोले एक दूसरे की ओर वेग v से आते हैं। तब गोलों के बीच की दूरी, x के फलन के रूप में वेग v का विचरण किस रूप में होता है?

Accepted Answer

चित्र से, tanθ=mgFe​​≃θया x3∝q2.........(i)
या x3/2∝q.........(ii)समय के सापेक्ष समीकरण (1) का अवकलन करने पर3x2dtdx​∝2qdtdq​ परन्तु dtdq​= स्थिरांकअत: x2(v)∝q समीकरण (2) से q का मान रखने पर x2(v)∝x3/2 या v∝x−1/2

Question 4

In the following diagram the work done in moving a point charge from point P to point A,B and C is respectively as WA′​WB​ and WC′​ then

Accepted Answer

According to the figure, there is no other charge. A stationary charge when moved in a space of no field, does not  apply any force. No work is doneWA​=WB​=WC​=0

Question 5

An object A has a charge of −2μC and the object B has a charge of +6μC. Which statement is true?

Accepted Answer

Vector from of Coulomb's law states that FAB​ and FBA​ are equal  and opposite.Since, r^AB​ and r^BA​ are unit vectors pointing in opposite direction, we haver^AB​=−r^BA​Therefore, equations of force is
Given byFAB​​=4πε0​1​r2q1​q2​​(r^AB​)=−4πε0​1​r2q1​q2​​r^BA​=−FBA​​

Question 6

.Three charges each of magnitude q are placed at the corners of an equilateral triangle, the electrostatic force on the charge placed at the center is (each side of triangle is L )

Accepted Answer

In the following figure since ∣∣​FA​∣∣​=∣∣​FB​∣∣​=∣∣​FC​∣∣​ and they are equally inclined with each other, so their resultant will be Zero

Question 7

When the distance between the charged particles is halved, the force between them becomes

Accepted Answer

F∝r21​; so when r is halved the force becomes four times

Question 8

If a conducting medium is placed between two charges, then the electric force between them will become.

Accepted Answer

When a conducting medium is placed between the two charges, then both charges combine in such a way that their combination behaves like a single charge. Electrical force acts only between two different charges. But, due to the absence of two different charges after the combination of charges with conducting medium,the force between them becomes zero.

Question 9

A uniformly charged thin spherical shell of radius R carries uniform surface charge density of σ per unit area. It is made of two hemispherical shells, held together by pressing them with force F (see figure). F is proportional to

Accepted Answer

Electrostatic repulsive force;Fele ​=(2ε0​σ2​)πR2;F=Fele ​=2ε0​σ2πR2​

Question 10

Two point charges +9e and +e are at 16 cm away from each other. Where should another charge q be placed between them so that the system remains in equilibrium

Accepted Answer

For equilibrium statex29×qc​​=(16−x)2q×e​ or, x29​=(16−x)21​ or, x3​=16−x1​or, 3(16−x)=xor, 48−3x=xor, 4x=48 or, x=12 (Ans) Now, from x the distance will be (16−12)cm=4 cm.

Question 11

Force between two identical charges placed at a distance of r in vacuum is F. Now a slab of dielectric of dielectric constant 4 is inserted between these two charges. If the thickness of the slab is r/2, then the force between the charges will become

Accepted Answer

From Coulomb's law the force (F) between two charges isF=4πε0​k1​r2q2​First case k=1F=4πε0​1​⋅r2q2​................(i)Second case k=4F′=4πε0​×41​r2q2​⇒F′F​=4F′=4F​

Question 12

Three charges −q1​+q2​ and −q3​ are placed as shown in figure. The x component of the force on −q1​ is proportional to

Accepted Answer

Force on −q1​ due to q2​ isF12​=b2kq1​q2​​ along X-axisForce on −q1​ due to −q3​ isF13​=a2kq1​q3​​ at ∠θ With negative direction of Y-axis.∴x component of force on −q1​ isFx​=F12​+F13​sinθ=kq1​[b2q2​​+a2q3​​sinθ]ie, Fx​∝[b2q2​​+a2q3​​sinθ]

Question 13

Three charges each of +1μC are placed at the corners of an equilateral triangle. If the force between any two charges be F, then the net force on either charge will be

Accepted Answer

Angle between two forces due to individual charges is equal and at 60∘∴R=F2+F2+2FFcoscos60∘​=F3​

Question 14

An α-particle of mass 6.4×10−27 kg and charge 3.2×10−19C is situated in a uniform electric field of 1.6×105Vm−1. The velocity of the particle at the end of 2×10−2 m path when it starts from rest is

Accepted Answer

Here, mass of α− particle ma​=6.4×10−27 kg
Charge of α− particle qa​=3.2×10−19C
Electric field E=1.6×105Vm−1
Force on α - particle in a uniform electric field E is F=qα​E=3.2×10−19×1.6×105 N
=3.2×1.6×10−14N
Acceleration of the α - particle isa=ma​F​=6.4×10−273.2×1.6×10−14​ ms−2

Given : u=0∴v2=2aS=6.4×10−272×3.2×1.6×10−14×2×10−2​

[As S=2×10−2 m ]
=3.2×1011=32×1010
or v=42​×105 ms−1

Question 15

A charge of 5C experiences a force of 5000N when it is kept in a uniform electric field. What is the potential difference between two points separated by a distance of 1 cm

Accepted Answer

Charge =5CForce =5000 NDistance =0.01 mV= ?t=qF orce ​=55000​=1000 Valso, E=drdv​⇒dv=E(dr)=1000100(0.01)​ mV=10 V

Question 16

Four charges arranged the at the corners of a square ABCD, as shown in the adjoining figure. The force on the charge kept at the centre O is

Accepted Answer

We put a unit positive charge at O. Resultant force due to the charge placed at A and C is zero and resultant charge due to B and D is towards D along the diagonal BD

Question 17

A total charge Q is broken in two parts Q1​ and Q2​ and they are placed at a distance R from each other. The maximum force of repulsion between them will occur, when

Accepted Answer

Q1​+Q2​=Q… (i) and  F=kr2Q1​Q2​​… (ii) 
From (i) and (ii)  F=r2kQ1​(Q−Q1​)​  
For F to be maximum dQ1​dF​=0⇒Q1​=Q2​=2Q​

Question 18

A charge Q.is divided in two parts Q−q. What is value of q for maximum force between them?

Accepted Answer

If r is the distance between q and (Q−q), the Coulomb force between them isF​=4πε0​1​⋅r2q(Q−q)​=4πε0​1​⋅r2(qQ−q2)​​For F to be maximum dF/dq=0ie dqd​[4πε0​1​(qQ−q2)]=0
i.e, Q−2q​=0(( as 4πε0​1​=0) or Qq​=21​⇒q=2Q​

Question 19

Two protons A and B are placed in space between plates of a parallel plate capacitor charged upto V volts (see fig.) Force on protons are FA​ and FB​ then

Accepted Answer

FA​=FB′​; because an uniform electric field produced between the plates

Question 20

Charge q2​ of mass m revolves around a stationary charge q1​ in a circular orbit of radius r. The orbital periodic time of q2​ would be

Accepted Answer

The force acting on the particle can be balanced by equating the electrostatic force by the centripetal force on the charge.4πε0​1​r2q1​q2​​T2T​=mrω2=T24π2mr2​=q1​q2​(4πε0​)r2(4π2mr)​=[kq1​q2​4π3mr2​]1/2​

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Solution for Class 12, Physics, Electric Charges and Fields

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