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Electrostatics NEET Questions — Free MCQ Practice

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In a region, the potential is represented by V(x,y,z)=6x-8xy-8y+6yz, where V is in volts and x,y,z are in meters. The electric force experienced by a charge of 2 coulomb situated at point (1,1,1) is

A.

6√5N

B.

30N

C.

24N

D.

4√35N

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Explanation

We know

F=qE ...(i)

E=-dV/dr

E_x= $δ$ V/ $δ$ x=6-8y

E_y= $δ$ V/ $δ$ y=-8x-8+6z ...(ii)

E_z=6y

Above values of E_x,Ey and E_z at (1,1,1) are

E_{x= $6 - 8 x (1) = - 2$}Ey=-8(1)-8+6(1)=-10

E_{z= $6 x 1 = 6$}

So, E_{net=√(-2)2+(10)2+(6)2}

_{=√4+100+36=√140 =>√35x4=2√35 N/C}

_{So, F=qEnet=2(2√35)=4√35N}

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Four point charges $- Q, - q, 2 q a n d 2 Q$ are placed, one at each corner of the square.The relation between Q and q for which the potential at the centre of the square is zero, is

A.

Q=-q

B.

Q=- $\frac{1}{q}$

C.

Q=q

D.

Q= $\frac{1}{q}$

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Explanation

If potential at centre is zero, then

$V_{1} + V_{2} + V_{3} + V_{4} = 0$

$- \frac{k Q}{r} - \frac{k q}{r} + \frac{k 2 Q}{r} + \frac{k 2 q}{r} = 0$

$- Q - q + 2 q + 2 Q = 0$

$Q = - q$

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Two metallic spheres of radii 1 cm and 3 cm

are given charges of -1 $\times 10^{- 2} C$ and $5 \times 10^{- 2} C$ ,

respectively. If these are connected by a conducting

wire, the final charge on the bigger sphere is

A.

$2 \times 10^{- 2} C$

B.

$3 \times 10^{- 2} C$

C.

$4 \times 10^{- 2} C$

D.

$1 \times 10^{- 2} C$

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Explanation

Charge flows from high potential to low

potential

$\frac{{KQ}_{1}}{3} = \frac{{KQ}_{2}}{1} Q_{1} = 3 Q_{2}$

Also, $Q_{1} + Q_{2} = 4 \times 10^{- 2} 4 Q_{2} = 4 \times 10^{- 2}$

$Q_{2} = 10^{- 2}$

and $Q_{1} = 3 \times 10^{- 2} C$

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A parallel plate condenser has a uniform electric

field E(V/m) in the space between the plates. If

the distance between the plates is d(m) and area

of each plate is $A (m^{2})$ , the energy (joule) stored

in the condenser is

A.

$\frac{1}{2} ε_{0} E^{2}$

B.

$ε_{0} EAd$

C.

$\frac{1}{2} ε_{0} E^{2} Ad$

D.

$E^{2} Ad / ε_{0}$

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Explanation

The energy stored in the condenser

$U = \frac{1}{2} {CV}^{2} U = \frac{1}{2} (\frac{{Aε}_{0}}{d}) (Ed)^{2} (so, C = \frac{{Aε}_{0}}{d} and V = Ed) U = \frac{1}{2} ε_{0} E^{2} Ad$

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A series combination of n₁ capacitors, each of value C₁, is charged by a source of potential difference 4V. When another parallel combination of n₂ capacitors, each of value C₂, is charged by a source of potential difference V, it has the same (total) energy stored in it, as the first combination has. The value of C₂, in terms of C₁, is then

A.

$\frac{2 C_{1}}{n_{1} n_{2}}$

B.

16 $\frac{n_{2}}{n_{1}}$ C₁

C.

2 $\frac{n_{2}}{n_{1}}$ C₁

D.

$\frac{16 C_{1}}{n_{1} {n_{2}}_{}}$

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Explanation

Case I. When the capacitors are joined in series

U_series = $\frac{1}{2}$ $\frac{C_{1}}{n_{1}}$ (4V)²

Case II. When the capacitors are joined in parallel

U_parallel= $\frac{1}{2}$ (n₂C₂)V²

Given, U_series= U_parallel

or $\frac{1}{2}$ $\frac{C_{1}}{n_{1}}$ (4V)² = $\frac{1}{2}$ (n₂C₂)V²

$\Rightarrow$ C₂ = $\frac{16 C_{1}}{n_{2} n_{1}}$

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Three concentric spherical shells have radii a, b and c (a<b<c) and have surface charge densities $σ, - σ$ and $σ$ respectively. If $V_{A}, V_{B}$ and $V_{C}$ denote the potential of the three shells, if c=a+b, we have

A.

$V_{C} = V_{A} \neq V_{B}$

B.

$V_{C} = V_{B} \neq V_{A}$

C.

$V_{C} \neq V_{B} \neq V_{A}$

D.

$V_{C} = V_{B} = V_{A}$

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Explanation

$Here, Potential at the surface of A : V_{A} = \frac{1}{4 {πε}_{0}} \frac{σ 4 {πa}^{2}}{a} - \frac{1}{4 {πε}_{0}} \frac{σ 4 {πb}^{2}}{b} + \frac{1}{4 {πε}_{0}} \frac{σ 4 {πc}^{2}}{c} = \frac{σ}{ε_{0}} (a - b + c) = \frac{σ}{ε_{0}} (2 a) (∵ c = a + b)$

$Potential at the surface of B : V_{B} = \frac{1}{4 {πε}_{0}} \cdot \frac{σ 4 {πa}^{2}}{b} - \frac{1}{4 {πε}_{0}} \frac{σ 4 {πb}^{2}}{b} + \frac{1}{4 {πε}_{0}} \frac{σ 4 {πc}^{2}}{c} = \frac{σ}{ε_{0}} (\frac{a^{2}}{b} - b + c) = \frac{σ}{ε_{0}} (\frac{a^{2}}{b} + a) (∵ c = a + b) and V_{C} = \frac{1}{4 {πε}_{0}} \cdot \frac{σ 4 {πa}^{2}}{c} - \frac{1}{4 {πε}_{0}} \frac{σ 4 {πb}^{2}}{c} + \frac{1}{4 {πε}_{0}} \frac{σ 4 {πc}^{2}}{c} = \frac{σ}{ε_{0}} (\frac{a^{2}}{c} - \frac{b^{2}}{c} + c) = \frac{σ}{ε_{0}} (\frac{a^{2} - b^{2} + c^{2}}{c}) = \frac{σ}{ε_{0}} (\frac{a^{2} - b^{2} + {(a + b)}^{2}}{c}) = \frac{σ}{ε_{0}} (2 a) (∵ c = a + b) Hence, V_{A} = V_{C} \neq V_{B}$

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The energy required to charge a parallel plate condenser of plate separation d and plate area of cross-section A such that the uniform electric field between the plates is E, is

A.

$\frac{1}{2} ε_{0} E^{2} / A d$

B.

$ε_{0} E^{2} / A d$

C.

$ε_{0} E^{2} A d$

D.

$\frac{1}{2} ε_{0} E^{2} A d$

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Explanation

Energy given by the cell

$E = \frac{1}{2} C V^{2}$

Here, C = capacitance of condenser = $\frac{A ε_{0}}{d}$

V = potential difference across the plates = Ed

Therefore, E = $\frac{1}{2} (\frac{A ε_{0}}{d}) {(E d)}^{2}$

= $\frac{1}{2} A ε_{0} E^{2} d$

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100 capacitors each having a capacity of 10 μF are connected in parallel and are charged by a potential difference of 100 kV. The energy stored in the capacitors and the cost of charging them, if electrical energy costs 108 paise per kWh, will be

A.

10⁷ joule and 300 paise

B.

5 × 10⁶ joule and 300 paise

C.

5 × 10⁶ joule and 150 paise

D.

10⁷ joule and 150 paise

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Explanation

Energy stored in the capacitor $= \frac{1}{2} C V^{2} \times 100$

$= \frac{1}{2} \times 10 \times 10^{^{- 6}} \times {(100 \times 10^{3})}^{2} \times 100 = 5 \times 10^{6} J$

Electric energy costs $= 108 P a i s e p e r k W H$ $= \frac{108 P a i s e}{3 .6 \times 10^{6} J}$

∴ Total cost of charging $= \frac{2 \times 5 \times 10^{6} \times 108}{3 .6 \times 10^{6}} = 300 P a i s e$

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A 10 μF capacitor and a 20 μF capacitor are connected in series across a 200 V supply line. The charged capacitors are then disconnected from the line and reconnected with their positive plates together and negative plates together and no external voltage is applied. What is the potential difference across each capacitor

A.

$\frac{400}{9} V$

B.

$\frac{800}{9} V$

C.

400 V

D.

200 V

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Explanation

Initially potential difference a cross each capacitor

$V_{1} = \frac{20}{(10 + 20)} \times 200 = \frac{400}{3} V$

and $V_{2} = \frac{10}{(10 + 20)} \times 200 = \frac{200}{3} V$

Finally common potential $V = \frac{C_{1} V_{1} + C_{2} V_{2}}{C_{1} + C_{2}}$

$V = \frac{10 \times \frac{400}{3} + 20 \times \frac{200}{3}}{(10 + 20)} = \frac{800}{9} V$

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Three capacitors of capacitance 3 μF, 10 μF and 15 μF are connected in series to a voltage source of 100V. The charge on 15 μF is

A.

50 μC

B.

100 μC

C.

200 μC

D.

280 μC

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Explanation

$\frac{1}{C_{e q}} = \frac{1}{3} + \frac{1}{10} + \frac{1}{15} \Rightarrow C_{e q} = 2 μ F$

Charge on each capacitor

Q = C_eq × V

$\Rightarrow 2 \times 100 = 200 μ C$

Electrostatics MCQs for NEET — Physics Questions with Answers

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