Question

14. The red dot in the sketch below represents a charged particle. The green dot represents a point on an equipotential surface. The blue dot represents a point on a different equipotential surface. If the potential at the blue dot is 60 v and the potential at the green dot is 100 v, in which direction is the E-field at the blue dot?

a. To the right
b. To the left
c. Toward the green dot
d. Away from the green dot

15. Based on the information given in question #14, what is the sign on the work done by the field when a small positive test charge moves from the blue dot to the green dot?

a. positive
b. negative
c. zero...due to the equipotentials shown

Answer 1

We know that E = dV/dr, if V increase with increase in r then dV/dr is positive and the electric should point towards the center. If V decrease with increase in r, then dV/dr is negative and the electric field must point away from the center. Since V is decreasing as r increase, the electric field must point away from the center and must point away from green dot. 15. Work done in general is given by W = q integral vector E. vector dr = -q integral dV = -q(V_2 - V_1) = -q(100 - 60) W = -40q Since q is positive, the work done is negative.