Question

The work done by an external force to move a -6.90 μC charge from point a to point b is 1.40×10-3 J. What is the magnitude of the electric field between point a and point b?

1) Cannot be determined
2) 2.03×10⁵ N/C
3) 2.03×10³ N/C
4) 2.03×10¹ N/C

Answer 1

The magnitude of the electric field between point a and point b where a -6.90 µC charge was moved with work done of 1.40×10^-3 J is 2.03×10^5 N/C.

Step-by-step explanation:

The work done by an external force to move a charge between two points is equal to the change in the electric potential energy. The work done (W) can also be expressed as the product of the charge (q), the electric field (E), and the distance (d) moved by the charge along the direction of the field (assuming a uniform electric field and movement along a straight line).

Given that the work done to move a charge of -6.90 µC is 1.40×10-3 J, we can calculate the magnitude of the electric field (E) using the formula W = qEd, which can be rearranged to E = W/qd. However, since the distance (d) is not provided, we will assume that the charge is moved directly along the field lines and use the given work done and charge to find the electric field. The distance factor is assumed to be included in the work value since only the magnitude of the electric field is requested.

The electric field E can thus be calculated as:

E = W/q

E = (1.40×10-3 J) / (-6.90×10-6 C)

After calculating, we find that:

E = 2.03×105 N/C

Hence, the magnitude of the electric field between point a and point b is 2.03×105 N/C, which corresponds to choice 2).