The work done by an electric field when moving a charge on an equipotential surface is zero
because there is no change in electric potential energy. Electric lines of force are always
perpendicular to equipotential surfaces, meaning that any movement along the surface does
not alter the electric potential, confirming that no work is done.
The Correct answer is: Both (A) and (R) are correct and (R) is the correct explanation of (A)
A point charge $ +q $ is placed at the origin. A second point charge $ +9q $ is placed at $ (d, 0, 0) $ in Cartesian coordinate system. The point in between them where the electric field vanishes is:
A small bob of mass 100 mg and charge +10 µC is connected to an insulating string of length 1 m. It is brought near to an infinitely long non-conducting sheet of charge density \( \sigma \) as shown in figure. If the string subtends an angle of 45° with the sheet at equilibrium, the charge density of sheet will be :
Consider two infinitely large plane parallel conducting plates as shown below. The plates are uniformly charged with a surface charge density \( +\sigma \) and \( -\sigma \). The force experienced by a point charge \( +q \) placed at the mid point between the plates will be:
Due to presence of an em-wave whose electric component is given by \( E = 100 \sin(\omega t - kx) \, NC^{-1} \), a cylinder of length 200 cm holds certain amount of em-energy inside it. If another cylinder of same length but half diameter than previous one holds same amount of em-energy, the magnitude of the electric field of the corresponding em-wave should be modified as:
Let \( \alpha, \beta \) be the roots of the equation \( x^2 - ax - b = 0 \) with \( \text{Im}(\alpha) < \text{Im}(\beta) \). Let \( P_n = \alpha^n - \beta^n \). If \[ P_3 = -5\sqrt{7}, \quad P_4 = -3\sqrt{7}, \quad P_5 = 11\sqrt{7}, \quad P_6 = 45\sqrt{7}, \] then \( |\alpha^4 + \beta^4| \) is equal to: