Chapter 24, Problem 12P

The two charges in Figure P24.12 are separated by a distance d = 2.00 cm, and Q = +5.00 nC. Find (a) the electric potential at A, (b) the electric potential at B, and (c) the electric potential difference between B and A.

Figure P24.12

To determine

The electric potential at point $A$.

Answer to Problem 12P

The electric potential at point $A$ is $5431.98\text{V}$.

Explanation of Solution

Given Info: Distance between point $A$ and $B$ is $2\text{cm}$, Charge at point 1 is $+5\text{nC}$, Charge at point 2 is $+10\text{nC}$.

The diagram for the given figure having point $A$ and $B$ is given below.

Figure (1)

Write the expression to find out electric potential at point $A$,

$V_{\text{A}}=V_{{\text{A}}_1}+V_{{\text{A}}_2}$ (1)

Here,

$V_{{\text{A}}_2}$ is electric potential at point $A$ due to charge at point 2.

$V_{{\text{A}}_1}$ is electric potential at point $A$ due to charge at point 1.

Formula to calculate electric potential at point $A$ due to charge at point 1 is,

$V_{{\text{A}}_1}=\frac{k_e{q}_1}{r_1}$

Formula to calculate electric potential at point $A$ due to charge at point 2 is,

$V_{{\text{A}}_2}=\frac{k_e{q}_2}{r_2}$

Substitute $\frac{k_e{q}_1}{r_1}$ for $V_{{\text{A}}_1}$ and $\frac{k_e{q}_2}{r_2}$ for $V_{{\text{A}}_2}$ in equation (I).

$V_{\text{A}}=\frac{k_e{q}_1}{r_1}+\frac{k_e{q}_2}{r_2}$ (2)

$k_e$ is constant.

$q_1$ is charge at any point.1.

$r_1$ is distance between point 1 and point $A$.

$q_2$ is charge at point 2.

$r_2$ is distance between point $A$ and point 2.

Write the formula of Pythagoras theorem to calculate diagonal of square.

$\begin{array}{c}{r}_2=\sqrt{r_1{}^2+r_1{}^2}\\ =\sqrt{2}{r}_1\end{array}$

Substitute $2\text{cm}$ for $r_1$ in above equation.

$r_2=2\sqrt{2}\text{cm}$

Substitute $9\times {10}^9{\text{Nm}}^2/{\text{C}}^2$ for $k_{\text{e}}$, $+5\text{nc}$ for $q_1\text{or}Q$, $2\text{cm}$ for $r_1$, $+10\text{nC}$ for $q_{\text{2}}\text{or}2Q$ and $2\sqrt{2}\text{cm}$ for $r_2$ in above equation to find electric potential at point $A$ in equation (2).

$\begin{array}{c}{V}_A=\left(\frac{\left(9\times {10}^9{\text{Nm}}^{\text{2}}/{\text{C}}^{\text{2}}\right)\times \left(5\text{nC}\left(\frac{1\text{C}}{{10}^9\text{nC}}\right)\right)}{2\text{cm}\left(\frac{{10}^{-2}\text{m}}{\text{cm}}\right)}\right)+\left(\frac{\left(9\times {10}^9{\text{Nm}}^{\text{2}}/{\text{C}}^{\text{2}}\right)\times \left(10\text{nC}\left(\frac{1\text{C}}{{10}^9\text{nC}}\right)\right)}{2\sqrt{2}\text{cm}\left(\frac{{10}^{-2}\text{m}}{\text{cm}}\right)}\right)\\ =2250\text{Nm}/\text{C}+3181.98\text{Nm}/\text{C}\\ =5431.98\text{Nm}/\text{C}\\ =5431.98\text{V}\end{array}$

Thus, electric potential at point $A$ is $5431.98\text{V}$.

Conclusion:

Therefore, electric potential at point $A$ due to charge at point 1 and charge at point is $5431.98\text{V}$.

To determine

The electric potential at point $B$.

Answer to Problem 12P

electric potential at point $B$ is $6091\text{V}$.

Explanation of Solution

Write the expression to find out electric potential at point $B$,

$V_{\text{B}}=V_{{\text{B}}_1}+V_{{\text{B}}_2}$ (3)

Here,

$V_{{\text{B}}_1}$ is electric potential at point $B$ due to charge at point 1.

$V_{{\text{B}}_2}$ is electric potential at point $B$ due to charge at point 2.

Formula to calculate electric potential at point $B$ due to charge at point 1 is,

$V_{{\text{B}}_1}=\frac{k_e{q}_1}{r_2}$

Formula to calculate electric potential at point $B$ due to charge at point 2 is,

$V_{{\text{B}}_2}=\frac{k_e{q}_2}{r_1}$

Substitute $\frac{k_e{q}_1}{r_2}$ for $V_{{\text{B}}_1}$ and $\frac{k_e{q}_2}{r_1}$ for $V_{{\text{B}}_2}$ in equation (3).

$V_{\text{B}}=\frac{k_e{q}_1}{r_2}+\frac{k_e{q}_2}{r_1}$ (4)

Here,

$k_e$ is constant.

$q_1$ is charge at any point.1.

$r_1$ is distance between point 2 and point $B$.

$q_2$ is charge at point 2.

$r_2$ is distance between point B and point 1.

Substitute $9\times {10}^9{\text{Nm}}^{\text{2}}/{\text{C}}^{\text{2}}$ for $k_e$, $+5\times {10}^{-9}\text{C}$ for $q_1\text{or}Q$, $+10\times {10}^{-9}\text{C}$ for $q_2\text{or}2Q$, $2\sqrt{2}\times {10}^{-2}\text{m}$ for $r_2$, $2\times {10}^{-2}\text{m}$ for $r_1$. in equation (4).

$\begin{array}{c}{V}_B=\left(\frac{\left(9\times {10}^9{\text{Nm}}^{\text{2}}/{\text{C}}^{\text{2}}\right)\times \left(+5\text{nC}\left(\frac{1\text{C}}{{10}^9\text{nC}}\right)\right)}{2\sqrt{2}\text{cm}\left(\frac{{10}^{-2}\text{m}}{\text{cm}}\right)}\right)+\left(\frac{\left(9\times {10}^9{\text{Nm}}^{\text{2}}/{\text{C}}^{\text{2}}\right)\times \left(+10\text{nC}\left(\frac{1\text{C}}{{10}^9\text{nC}}\right)\right)}{2\text{cm}\left(\frac{{10}^{-2}\text{m}}{\text{cm}}\right)}\right)\\ =1591\text{Nm}/\text{C}+4500\text{Nm}/\text{C}\\ =6091\text{Nm}/\text{C}\\ =6091\text{V}\end{array}$

Thus, electric potential at point B is $6091\text{V}$.

Conclusion:

Therefore, the electric potential at point B due to charge at point 1 and charge at point 2 is $6091\text{V}$.

To determine

The electric potential difference between $B$ and $A$.

Answer to Problem 12P

Potential difference between point $A$ and $B$ is $659.02\text{V}$.

Explanation of Solution

Write the expression to find out electric potential difference between $A$ and $B$.

$\Delta V=V_{\text{B}}-V_{\text{A}}$

Here,

$\Delta V$ is electric potential difference between point $A$ and point $B$.

$V_{\text{B}}$ is electric potential at point $B$.

$V_{\text{A}}$ is electric potential at point $A$.

Substitute $6091\text{V}$ for ${\text{V}}_{\text{B}}$ and $5431.98\text{V}$ for ${\text{V}}_{\text{A}}$ in above equation to find out the value of electric potential difference,

$\begin{array}{c}\Delta V=\left(6091\text{V}\right)-\left(5431.98\text{V}\right)\\ =659.02\text{V}\end{array}$

Hence, electric potential difference between point $A$ and $B$ is $659.02\text{V}$.

Conclusion:

Therefore, electric potential difference between point $A$ and $B$ is $659.02\text{V}$.