Assume that an electric field E = 30x^2 , i N/C exists in space. If V₀ is the potential at the origin and V_A is the potential at x = 2 m, then the potential difference V_A - V₀ is

Step-by-step solution

The relation between electric potential and electric field is given by dV = - E d r . Since the electric field is only in the x-direction, we have dV = -E_x dx . Integrating both sides from the origin ( x = 0 ) to x = 2 m: _ V₀ ^ V_A dV = - ₀² 30x^2 dx V_A - V₀ = -30 [ x^3 3 ]₀² V_A - V₀ = -10 (2^3 - 0^3) V_A - V₀ = -80 V. Answer: -80 V