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Electrostatics NEET Questions — Free MCQ Practice

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An elementary particle of mass m and charge +e is projected with velocity v at a much more massive particle of charge Ze, where Z > 0. What is the closest possible approach of the incident particle ?

A.

$\frac{Z e^{2}}{2 π ε_{0} m v^{2}}$

B.

$\frac{Z e}{4 π ε_{0} m v^{2}}$

C.

$\frac{Z e^{2}}{8 π ε_{0} m v^{2}}$

D.

$\frac{Z e}{8 π ε_{0} m v^{2}}$

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Explanation

Suppose distance of closest approach is r, and according to energy conservation applied for elementary charge.

Energy at the time of projection = Energy at the distance of closest approach

⇒ $\frac{1}{2} m v^{2} = \frac{1}{4 π ε_{0}} . \frac{(Z e) . e}{r} \Rightarrow r = \frac{Z e^{2}}{2 π ε_{0} m v^{2}}$

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A solid conducting sphere having a charge Q is surrounded by an uncharged concentric conducting hollow spherical shell. Let the potential difference between the surface of the solid sphere and that of the outer surface of the hollow shell be V. If the shell is now given a charge of –3Q, the new potential difference between the same two surfaces is

A.

V

B.

2V

C.

4V

D.

–2V

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Explanation

In case of a charged conducting sphere

$V_{inside} = V_{centre} = V_{surface} = \frac{1}{4 π ε_{o}} . \frac{q}{R}$ , $V_{outside} = \frac{1}{4 π ε_{0}} . \frac{q}{r}$

If a and b are the radii of sphere and spherical shell respectively, then potential at their surface will be

$V_{sphere} = \frac{1}{4 π ε_{0}} . \frac{Q}{a}$ and $V_{shell} = \frac{1}{4 π ε_{0}} . \frac{Q}{b}$

∴ $V = V_{sphere} - V_{shell} = \frac{1}{4 π ε_{0}} . [\frac{Q}{a} - \frac{Q}{b}]$

Now when the shell is given charge (–3Q), then the potential will be

$V'_{sphere} = \frac{1}{4 π ε_{0}} [\frac{Q}{a} + \frac{(- 3 Q)}{b}], V'_{shell} = \frac{1}{4 π ε_{0}} [\frac{Q}{b} + \frac{(- 3 Q)}{b}]$

∴ $V'_{sphere} - V'_{shell} = \frac{1}{4 π ε_{0}} [\frac{Q}{a} - \frac{Q}{b}] = V$

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A piece of cloud is having area 25 × 10⁶ m² and electric potential of 10⁵ volts. If the height of cloud is 0.75 km, then energy of electric field between earth and cloud will be

A.

250 J

B.

759 J

C.

1225 J

D.

1475 J

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Explanation

Energy $= \frac{1}{2} ε_{0} E^{2} \times (A \times d) = \frac{1}{2} ε_{0} (\frac{V^{2}}{d^{2}}) A d$

$= \frac{1}{2} \times \frac{8 .85 \times 10^{- 12} \times {(10^{5})}^{2} \times 25 \times 10^{6}}{0 .75 \times 10^{3}} = 1475 J$

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Capacitance of a capacitor made by a thin metal foil is 2 μF. If the foil is folded with paper of thickness 0.15 mm, dielectric constant of paper is 2.5 and width of paper is 400 mm, then length of foil will be

A.

0.34 m

B.

1.33 m

C.

13.4 m

D.

33.9 m

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Explanation

If length of the foil is l then $C = \frac{k ε_{0} (l \times b)}{d}$

⇒ $2 \times 10^{- 6} = \frac{2 .5 \times 8 .85 \times 10^{- 12} (l \times 400 \times 10^{- 3})}{0 .15 \times 10^{- 3}}$

⇒ l = 33.9 m

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A parallel plate capacitor is connected to a battery. The plates are pulled apart with a uniform speed. If x is the separation between the plates, the time rate of change of electrostatic energy of the capacitor is proportional to:

A.

x^–2

B.

x

C.

x^–1

D.

x²

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Explanation

$U = \frac{1}{2} C V^{2} = \frac{1}{2} (\frac{ε_{0} A}{x}) V^{2}$

∴ $\frac{d U}{d t} = \frac{1}{2} ε_{0} A V^{2} (- \frac{1}{x^{2}} \frac{d x}{d t})$

⇒ $\frac{d U}{d t} \propto x^{- 2}$

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To form a composite 16 μF, 1000 V capacitor from a supply of identical capacitors marked 8 μF, 250 V, we require a minimum number of capacitors

A.

40

B.

32

C.

8

D.

2

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Explanation

To form a composite 16 Î¼F, 1000 V capacitor from identical 8 Î¼F, 250 V capacitors, we need to connect them in parallel. The equivalent capacitance increases, while the maximum voltage rating remains the same as the lowest voltage rating of individual capacitors. Since 16/8 = 2, we need to connect 2 capacitors in parallel, which requires 32 capacitors.

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Two condensers of capacities 2C and C are joined in parallel and charged upto potential V. The battery is removed and the condenser of capacity C is filled completely with a medium of dielectric constant K. The p.d. across the capacitors will now be

A.

$\frac{3 V}{K + 2}$

B.

$\frac{3 V}{K}$

C.

$\frac{V}{K + 2}$

D.

$\frac{V}{K}$

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Explanation

$q_{1} = 2 C V, q_{2} = C V$

Now condenser of capacity C is filled with dielectric K, therefore C₂ = KC

As charge is conserved

∴ $q_{1} + q_{2} = (C_{2}^{'} + 2 C) V'$

⇒ $V' = \frac{3 C V}{(K + 2) C} = \frac{3 V}{K + 2}$

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A parallel plate capacitor of capacitance C is connected to a battery and is charged to a potential difference V. Another capacitor of capacitance 2C is connected to another battery and is charged to potential difference 2V. The charging batteries are now disconnected and the capacitors are connected in parallel to each other in such a way that the positive terminal of one is connected to the negative terminal of the other. The final energy of the configuration is

A.

Zero

B.

$\frac{25 C V^{2}}{6}$

C.

$\frac{3 C V^{2}}{2}$

D.

$\frac{9 C V^{2}}{2}$

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Explanation

Total charge $= (2 C) (2 V) + (C) (- V) = 3 C V$

∴ Common potential $= \frac{3 C V}{3 C} = V$

∴ Energy $= \frac{1}{2} (3 C) {(V)}^{2} = \frac{3}{2} C V^{2}$

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Condenser A has a capacity of 15 μF when it is filled with a medium of dielectric constant 15. Another condenser B has a capacity of 1 μF with air between the plates. Both are charged separately by a battery of 100 V. After charging, both are connected in parallel without the battery and the dielectric medium being removed. The common potential now is

A.

400 V

B.

800 V

C.

1200 V

D.

1600 V

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Explanation

Charge on capacitor A is given by $Q_{1} = 15 \times 10^{- 6} \times 100 = 15 \times 10^{- 4} C$

Charge on capacitor B is given by $Q_{2} = 1 \times 10^{- 6} \times 100 = 10^{- 4} C$

Capacity of capacitor A after removing dielectric $= \frac{15 \times 10^{- 6}}{15} = 1 μ F$

Now when both capacitors are connected in parallel their equivalent capacitance will be C_eq = 1 + 1 = 2 μF

So common potential $= \frac{(15 \times 10^{- 4}) + (1 \times 10^{- 4})}{2 \times 10^{- 6}} = 800 V .$

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A capacitor of capacitance C₁ = 1 μF can withstand maximum voltage V₁ = 6kV (kilo-volt) and another capacitor of capacitance C₂ = 3 μF can withstand maximum voltage V₂ = 4 kV. When the two capacitors are connected in series, the combined system can withstand a maximum voltage of

A.

4 kV

B.

6 kV

C.

8 kV

D.

10 kV

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Explanation

As Q = CV, (Q₁)_max=10^–6 × 6 × 10³ = 6mC

While (Q₂)_max= 3 × 10^–6 × 4 × 10³ = 12mC

However in series charge is same so maximum charge on C₂ will also be 6 mC (and not 12 mC) and potential difference across it V₂ = 6mC/3 μF = 2KV and as in series V =V₁ + V₂ so V_max= 6KV + 2KV = 8KV

Electrostatics MCQs for NEET — Physics Questions with Answers

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