The electric field in a region is given by E = 6x^2 i + y^2 j (in V/m ), where x and y are in meters. If the electric potential at the origin (0,0) is 40 V higher than the electric potential at the point (2,2) m , the value of the constant (in V/m ^3 ) is

Step-by-step solution

The relationship between electric potential difference and electric field is given by: V_A - V_B = _ A ^ B E d r Here, we are given V(0,0) - V(2,2) = 40 V . We can set up the integral from the origin (0,0) to the point (2,2) : V(0,0) - V(2,2) = _ (0,0) ^ (2,2) (6x^2 i + y^2 j ) (dx i + dy j ) 40 = ₀² 6x^2 dx + ₀² y^2 dy Evaluating the integrals: 40 = [ 6x^3 3 ]₀² + [ y^3 3 ]₀² 40 = [ 2x^3 ]₀² + [ y^3 3 ]₀² 40 = 2(8) + (8) 3 40 = 16 + 8 3 Subtracting 16 from both sides: 24 = 8 3 Solving for : = 24 3 8 = 9 Answer: 9