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Electrostatics NEET Questions — Free MCQ Practice

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The electric potential at a point in space due to charge Q is Q $\times 10^{12}$ V. The value of an electric field at that point will be

A.

$\frac{Q}{9} \times 10^{12} \frac{N}{C}$

B.

$\frac{Q}{9} \times 10^{15} \frac{N}{C}$

C.

$\frac{Q}{9} \times 10^{9} \frac{N}{C}$

D.

$Q \times 10^{15} \frac{N}{C}$

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Explanation

$\begin{aligned} (2) \\ V = \frac{KQ}{r} \\ Q \times 10^{12} = \frac{9 \times 10^{9} Q}{r} \\ r = 9 \times 10^{- 3} \\ E = \frac{KQ}{r^{2}} = \frac{9 \times 10^{9} Q}{9 \times 9 \times 10^{- 6}} \\ E = \frac{Q \times 10^{15}}{9} N / C \end{aligned}$

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The electric potential at a point at distance 'r' from a short dipole is proportional to

A.

$r^{2}$

B.

$r^{- 1}$

C.

$r^{- 2}$

D.

$r^{1}$

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Explanation

$V = \frac{K P}{r^{2}}$

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A hollow charged metal spherical shell has radius R. If the potential difference between its surface and a point at a distance 3R from the center is V, then the value of electric field intensity at a point at distance 4R from the center is

A.

$\frac{3 V}{19 R}$

B.

$\frac{V}{6 R}$

C.

$\frac{3 V}{32 R}$

D.

$\frac{3 V}{16 R}$

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Explanation

3.

$V = \frac{Kq}{R} - \frac{Kq}{3 R} = \frac{2 Kq}{3 R} E = \frac{Kq}{16 R^{2}} = \frac{3 RV}{2 \times 16 R^{2}} E = \frac{3 V}{32 R}$

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A metallic sphere is given some charge. Electric energy is stored

A.

Only inside the sphere

B.

Only outside the sphere

C.

Both inside and outside

D.

Neither inside nor outside

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Explanation

Charge always resides on the outer surface.

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Capacitors $C_{1} = 10 μ F a n d C_{2} = 30 μ F$ are connected in series across a source of emf 20KV. The potential difference across $C_{1}$ will be

A.

5 KV

B.

15 KV

C.

10 KV

D.

20 KV

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Explanation

In a series combination of capacitors, the potential difference across each capacitor is inversely proportional to its capacitance. Since C2 = 30Î¼F is thrice C1 = 10Î¼F, the potential drop across C1 will be thrice that across C2. With a total potential of 20KV, the drop across C1 is 15KV.

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Two metallic spheres of radii 2cm and 3cm are given charges 6mC and 4mC respectively. The final charge on the smaller sphere will be if they are connected by a conducting wire

A.

4mC

B.

6mC

C.

5mC

D.

10mC

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Explanation

Finally, both spheres need to have equal potential. Let final charges be $q_{1} a n d q_{2}$

$\frac{k q_{1}}{2} = \frac{k q_{2}}{3} \frac{q_{1}}{q_{2}} = \frac{2}{3} A l s o, q_{1} + q_{2} = 10 S o l v i n g, q_{1} = 4 m C$

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When a proton at rest is accelerated by a potential difference V, its speed is found to be v. The speed of an $α$ particle when accelerated by the same potential difference from rest will be

A.

v

B.

$\frac{v}{\sqrt{2}}$

C.

$v \sqrt{2}$

D.

2v

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Explanation

$q V = \frac{1}{2} m v^{2}$

assuming charge on Proton as q and mass as m, above is the equation based on conservation of energy (decrease in potential energy is equal to the increase in kinetic energy here)

therefore, $v_{p r o t o n}$ = $\sqrt{\frac{2 q V}{m}}$

As α particle has 2 protons and 2 neutrons, so charge = 2q and mass = 4m and the equation becomes

^{$2 q V =$ $\frac{1}{2} \times 4 m \times {v_{α}}^{2}_{^{}}$}

So, $v_{α}$ = √(qV/m)

Therefore, $v_{α}$ = $\frac{v_{p r o t o n}}{\sqrt{2}}$

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Four equal charges Q are placed at the four corners of a square of each side is ‘a’. Work done in removing a charge – Q from its centre to infinity is

A.

0

B.

$\frac{\sqrt{2} Q^{2}}{4 π ε_{0} a}$

C.

$\frac{\sqrt{2} Q^{2}}{π ε_{0} a}$

D.

$\frac{Q^{2}}{2 π ε_{0} a}$

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Angle between equipotential surface and lines of force is

A.

Zero

B.

180°

C.

90°

D.

45°

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Explanation

Lines of force is perpendicular to the equipotential surface. Hence angle = 90°

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Two spheres of radius a and b respectively are charged and joined by a wire. The ratio of electric field of the spheres is

A.

a/b

B.

b/a

C.

a²/b²

D.

b²/a²

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Explanation

Joined by a wire means they are at the same potential. For same potential $\frac{k Q_{1}}{a_{1}} = \frac{k Q_{2}}{a_{2}}$

⇒ $\frac{Q_{1}}{Q_{2}} = \frac{a}{b}$

Further, the electric field at the surface of the sphere having radius R and charge Q is $\frac{k Q}{R^{2}} .$

∴ $\frac{E_{1}}{E_{2}} = \frac{k Q_{1} / a^{2}}{k Q_{2} / b_{2}} = \frac{Q_{1}}{Q_{2}} \times \frac{b^{2}}{a^{2}} = \frac{b}{a}$

Electrostatics MCQs for NEET — Physics Questions with Answers

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