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Electrostatics MCQs for NEET — Physics Questions with Answers

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An oil drop of 12 excess electrons is held stationary under a constant electric field of $ 2.55 \times 10^4 Vm^{–1} $ . If the density of the oil is $ 1.26 gm/cm^3$ then the radius of the drop is

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Explanation

As the drop is stationary , weight of drop = force due to electric field $ { 4 \over 3 } \pi r^3 \rho g = neE $ So, $ r^3 = { 3neE \over 4 \pi \rho g } $ now find out r

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Two points are at distances a and b (a < b) from a long string of charge per unit length $ \lambda $ The potential difference between the points in proportional to

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Explanation

$ E = - { dv \over dr } $ $ dv = { - \lambda \over 2 \pi \varepsilon_o r } $ $ { \lambda \over 2 \pi \varepsilon_o r } = { dv \over dr } $ $ \int _{va} ^ { vb } dv = -{ \lambda \over 2 \pi \varepsilon_0 } \int _a^b { 1 \over r } dr $

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Two Points P and Q are maintained at the Potentials of 10 v and –4 v, respectively. The work done in moving 100 electrons from P to Q is

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Explanation

$ W = 100 e (-4 -10 ) = -1400 ev = -1400 (-1.6 \times 10^{-19} ) J = 2.24 \times 10^{-16} J $

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Charges of $ + {10 \over 3 } \times 10^{-9} C $ are placed at each of the four corners of a square of side 8cm. The potential at the intersection of the diagonals is ....

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Three charges 2q, –q, –q are located at the vertices of an equilateral triangle. At the centre of the triangle.

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Explanation

In an equilateral triangle with charges 2q, -q, and -q at the vertices, the electric field at the center will not be zero because the configuration is asymmetric in terms of the magnitudes of charges. However, the potential at the center will be zero as the contributions from all three charges will cancel out each other. Hence, the field is non-zero and the potential is zero.

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Three particles, each having a charge of 10 c are placed at the corners of an equilateral triangle of side 10 cm. The electrostatic potential energy of the system is $ ( given { 1 \over 4 \pi \varepsilon _0 } = 9 \times 10^9 N.m^2 /c^2 )$

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Explanation

For pair of charge $ U = { 1 \over 4 } \pi \epsilon_0 \left[ { 10 \times 10^{-6} \times 10 \times 10^{-6} \over { 10 /100 } } \right] + \left[ { 10 \times 10^{-6} \times 10 \times 10^{-6} \over { 10 /100 } } \right] + \left[ { 10 \times 10^{-6} \times 10 \times 10^{-6} \over { 10 /100 } } \right] $ $ = { 3 \times 9 \times 10^ 9 \times 100 \times 10^{-2} \times 100 \over 10 } = 27 J $

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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 ...

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Two charged spheres of radii $R_1$ and $R_2$ having equal surface charge density. The ratio of their potential is …

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Explanation

For two spheres with radii $R_1$ and $R_2$ having equal surface charge density, the potential of a sphere is directly proportional to its radius. Therefore, the ratio of their potentials will be $\frac{R_1}{R_2}$.

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Two equal charges q are placed at a distance of 2a and a third charge -2q is placed at the midpoint. The potential energy of the system is ....

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Explanation

$ U system = { 1 \over 4 \pi \epsilon _o } { (q) (-2q ) \over q } + { 1 \over 4 \pi \epsilon_o } { (-2q) (q) \over a } + { 1 \over 4 \pi \varepsilon_o } { (q) (q) \over (2a) } = { -7 q^2 \over 8 \pi \varepsilon_o a }$

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Two point charges 100 c and 5 c are placed at points A and B respectively with AB = 40 cm. The work done by external force in displacing the charge 5 c from B to C where BC = 30 cm, angle $ ABC = { \pi \over 2 } and { 1 \over 4 \pi \varepsilon _0 } = 9 \times 10^9 Nm^2 / c^2 $

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