Question

PART A

What is the work done by the electric force to move a 1 C charge from A to B?

Express your answer in joules.

PART B

What is the work done by the electric force to move a 1 C charge from A to D?

Express your answer in joules.

Part C

The magnitude of the electric field at point C is

Electric Fields and Equipotential Surfaces The dashed lines in the diagram represent cross sections of equipotential surfaces drawn in 1 V increments. (Figure 1) | Figure 1 of 1 2 V -2 V 、、 -3 V

	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.

Answer 1

Concepts and reason

The concepts required to solve this problem are the electric field, potential difference between the two points, work done and the electrical potential energy.

Initially, obtain the expression for work done by the electric force to move a charge from one-point to another point B by using the expressions for potential difference between the two points work done.

After that, calculate the work done by the electric force to move a 1 C charge from one-point A to another point B and later, calculate the work done by the electric force to move a 1 C charge from one-point A to another point D by using the expression for work done by the electric force to move a charge from one-point to another point.

Finally, obtain the magnitude of the electric field at point C by using the electric field and provided figure.

Fundamentals

The potential difference between the two points is given as follows:

$\Delta V = {V_2} - {V_1}$

Here, ${V_1}\;{\rm{and }}{V_2}$ are the potentials at two points.

The work done is equal to negative gradient of potential. It is given by,

$W = - \Delta U$

The potential difference between the two points is equal to the potential energy between two points per unit charge. It is given by,

$\Delta V = \frac{{\Delta U}}{q}$

Here, q is the charge.

The electric field is given by,

$E = \frac{V}{d}$

Here, $d$ is the distance and V is the potential.

The potential difference between the two points is given as follows:

$\Delta V = {V_2} - {V_1}$

Here, ${V_1}\;{\rm{and }}{V_2}$ are the potentials at two points.

The potential difference between the two points is equal to the potential energy between two points per unit charge. It is given by,

$\Delta V = \frac{{\Delta U}}{q}$

Re-arrange the above expression as follows:

$\Delta U = q\left( {\Delta V} \right)$

Substitute ${V_2} - {V_1}$ for $\Delta V$ in the above expression.

$\Delta U = q\left( {{V_2} - {V_1}} \right)$

Therefore, the work done by the electric force to move a charge from one-point to another point is $W = q\left( {{V_1} - {V_2}} \right)$ .

The work done by the electric force to move a charge from one-point to another point is $W = q\left( {{V_1} - {V_2}} \right)$ .

Substitute ${W_{{\rm{AB}}}}$ for W, $1\;{\rm{C}}$ for q, $1\;{\rm{V}}$ for ${V_1}\;{\rm{and }}{V_2}$ in the above equation.

$\begin{array}{c}\\{W_{{\rm{AB}}}} = \left( {1\;{\rm{C}}} \right)\left( {1\;{\rm{V}} - 1\;{\rm{V}}} \right)\\\\ = 0\;{\rm{J}}\\\end{array}$

Therefore, the work done by the electric force to move a 1 C charge from A to B is $0\;{\rm{J}}$ .

The work done by the electric force to move a charge from one-point to another point is $W = q\left( {{V_1} - {V_2}} \right)$ .

Substitute ${W_{{\rm{AD}}}}$ for W, $1\;{\rm{C}}$ for q, $1\;{\rm{V}}$ for ${V_1}\;{\rm{and}}$ 0 V for ${V_2}$ in the above equation.

$\begin{array}{c}\\{W_{{\rm{AD}}}} = \left( {1\;{\rm{C}}} \right)\left( {1\;{\rm{V}} - 0\;{\rm{V}}} \right)\\\\ = 1\;{\rm{J}}\\\end{array}$

Therefore, the work done by the electric force to move a 1 C charge from A to D is $1\;{\rm{J}}$ .

The electric field is given by,

$E = \frac{V}{d}$

When the electric field lines are closer then the magnitude of the electric field strength will be more. When the electric field lines are far apart then the magnitude of the electric field strength will be less.

From the provided figure, the Point C is having the electric field lines which are closer compared to electric field lines at point B.

Hence, the magnitude of the electric field at point C is greater than the magnitude of electric field at point B.

Ans:

The work done by the electric force to move a 1 C charge from A to B is $0\;{\rm{J}}$ .

The work done by the electric force to move a 1 C charge from A to D is $1\;{\rm{J}}$ .

The magnitude of the electric field at point C is greater than the magnitude of electric field at point B.