An electric field of 20 N C ⁻¹ is present along the x -axis in space. Calculate the potential difference V_B - V_A when the points A and B are given by: (a) A = (0, 0) ; B = (4 m , 2 m ) (b) A = (4 m , 2 m ) ; B = (6 m , 5 m ) (c) A = (0, 0) ; B = (6 m , 5 m )

Step-by-step solution

The relation between potential difference and electric field is given by V_B - V_A = - _ A ^ B E d r . Since the electric field is uniform and directed along the x -axis, E = 20 i N C ⁻¹ . The potential difference becomes V_B - V_A = -E_x (x_B - x_A) = -20 (x_B - x_A) . For case (a): x_A = 0 m , x_B = 4 m V_B - V_A = -20(4 - 0) = -80 V For case (b): x_A = 4 m , x_B = 6 m V_B - V_A = -20(6 - 4) = -40 V For case (c): x_A = 0 m , x_B = 6 m V_B - V_A = -20(6 - 0) = -120 V