Question

the dashed lines in the diagram represent cross - sections of equipotential surfaces drawn in 1 v increments. (figure 1) what is the work done by the electric force to move a 1 c charge from a to d? express your answer in joules. view available hint(s) part c the magnitude of the electric field at point c is greater than the magnitude of the electric field at point b. less than the magnitude of the electric field at point b. equal to the magnitude of the electric field at point b. unknown because the value of the electric potential at point c is unknown. view available hint(s)

Explanation:

Step1: Recall the relationship between electric - field and equipotential lines

The electric - field magnitude $E$ is related to the potential difference $\Delta V$ and the distance $d$ between equipotential lines by $E=\frac{\Delta V}{d}$. The closer the equipotential lines, the greater the electric - field magnitude.

Step2: Analyze the spacing of equipotential lines at points C and B

At point C, the equipotential lines are closer together than at point B. Since $\Delta V$ is the same (1 - V increments for all equipotential lines in the diagram), and $E=\frac{\Delta V}{d}$, when $d$ (the distance between equipotential lines) is smaller, $E$ is larger. So the magnitude of the electric field at point C is greater than the magnitude of the electric field at point B.

Step3: Calculate the work done in moving a charge

The work done $W$ by an electric force in moving a charge $q$ between two points with a potential difference $\Delta V$ is given by $W = q\Delta V$. The potential at point A is $1\ V$ and the potential at point D is $- 2\ V$. So $\Delta V=V_D - V_A=-2 - 1=-3\ V$. Given $q = 1\ C$, then $W=q\Delta V=(1\ C)\times(- 3\ V)=-3\ J$. The magnitude of the work done is $3\ J$.

Answer:

Part C: greater than the magnitude of the electric field at point B
Work done: 3 J