Physics MCQs for NEET — Practice Questions with Answers

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.

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 $

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 $\frac{R_1}{R_2}$.

$ 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 }$

$ E = {-dv \over dx } = { -d \over dx } ( 5 x^2 + 10x -9 ) = -10x -10 $ $ \therefore E_(x-1) = - 10 \times 1 -10 = -20 { v \over m} $

Energy density $ ue = { 1 / 2 } \varepsilon_o E^2 = {1/2 } 8.86 \times 10^{-12 } \times ( {v \over r} ) ^2 = 2.83 J/m^3 $

Potential inside the Sphere will be same as that on i ts Surface i .e. $v = V_{surface} ={q / 10} volt$ $ Vout = { q /15 } volt $ $ \therefore { vout / v } = { 2 /3 } \Rightarrow V_{out} = { 2 /3 } V $

$ by using W = Q ( \bar E. |bar { \triangle r } ) $ $ W = Q [ ( e_1 \hat i + e_2 \hat j + e_3 \hat k ) . (a \hat i + b \hat j ) ] = Q (e_1 a + e_2 b ) $