A ball has a charge of -5 nC moves frompoint A to point B. The work done by the

Question

A ball has a charge of -5 nC moves frompoint A to point B. The work done by the electric field during themove is 1.7 mJ.
(a) What is the change in electric potential energy of the ball? Enter anumber. 1mJ
(b) What is the difference in voltage between points B and A; V(B)- V(A)? Enter anumber. 2V
(c) At which point does the ball have a higher electric potentialenergy? 3---Select---Point APoint BThey have the same voltageThere is not enough information
(d) Which point has the higher voltage? Enter anumber. Enter anumber.

Explanation / Answer

chrage q = -5 nC The work done by the electric field during the move is W =1.7 mJ.
(a) The change in electric potential energy of the ball U = -W = -1.7 mJ
(b) The difference in voltage between points B and A; V(B) -V(A) = W / q                                                                                                      = 1.7 * 10^ -3 J / ( -5 * 10 ^ -9 C )                                                                                                      = - 0.34* 10^ 6 Volts (c). At point A (d). At point A
(b) The difference in voltage between points B and A; V(B) -V(A) = W / q                                                                                                      = 1.7 * 10^ -3 J / ( -5 * 10 ^ -9 C )                                                                                                      = - 0.34* 10^ 6 Volts (c). At point A (d). At point A

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