CALC? The electric potential V in a region of space is given by where A is a constant. (a) Derive an expression for the electric field at any point in this region. (b) The work done by the field when a 1.50-µC test charge moves from the point (x, y, z) = (0, 0, 0.250 m) to the origin is measured to be 6.00 X 10-5 J. Determine A. (c) Determine the electric field at the point (0, 0, 0.250 m). (d) Show that in every plane parallel to the xz-plane the equipotential contours are circles. (e) What is the radius of the equipotential contour corresponding to V = 1280 V and y = 2.00 m?
Introduction First we have to calculate the electric field as a function of position. Then we have to calculate the constant A from the given work done. Then we have to calculate the electric field at the given point. Then we have to show that the equipotential contour parallel to xz plane are circle. And then we have to find out the radius of the equipotential couture at xz plane at y = 2.00 m Step 1 The potential is given by V (x,y,z) = A(x 3y + z )2 Now the electric field is given by 2 2 2 E(x,y,z) = V (x,y,z) = A(x 3y + z ) ˆ ˆ ˆ 2 2 2 = ( xi + y j + zk )(x 3y + z ) = 2A(xi 3yj + zk) ˆ ………………….(1) So the above equation represents the electric field as a function of x,y,z . Step 2 Now the work done in the electric field is given by W = q(V (x ,y ,z ) V (x ,y ,z ) = qA[(x 2 3y 2+ z ) (x 2 3y 2 + z )] 2 2 2 1 1 1 2 2 2 1 1 1 Here we have (x1,y 1z 1 = (0,0,0.250 m) and (x 2y 2z 2 = (0,0,0) 6 also q = 1.50 C = 1.50 × 10 C and W = 6.00 × 10 5 J So using the above values we have 5 6 2 6 × 10 J = (1.50 × 10 C)A( 0.25 m) 6×105J 2 A = (1.50×106C)(0.25 m) = 640 J/C.m ……………(2) Step 3 The electric field at point (0,0,0.250 m) is given by ˆ ˆ ˆ 2 ˆ ˆ E(0,0,0.25 m) = 2A(xi 3yj + zk) = 2(640 J/C.m )(0.250 m)k = ( 320 V /m)k Hence the electric field at the point (0,0,0.250 m) is ( 320 V /m)k .