Question

(II) Point a is 62 cm north of a \( - {\bf{3}}{\bf{.8}}\;{\bf{\mu C}}\) point charge, and point b is 88 cm west of the charge (Fig. 17–40). Determine (a) \({{\bf{V}}_{\bf{b}}} - {{\bf{V}}_{\bf{a}}}\) and (b) \({{\bf{\vec E}}_{\bf{b}}} - {{\bf{\vec E}}_{\bf{a}}}\) (magnitude and direction).

FIGURE 17–40 Problem 27.

Short answer

(a) The value of \({V_{\rm{b}}} - {V_{\rm{a}}}\)is \(1.6 \times {10^4}\;{\rm{V}}\).

(b) The magnitude of the electric field is \(9.9 \times {10^4}\;{{\rm{V}} \mathord{\left/{\vphantom {{\rm{V}} {\rm{m}}}} \right.} {\rm{m}}}\) and direction is \(64^\circ \) above the positive x-axis

Step-by-step solution

Step 1:Understanding of electric potential energy

The value of the electric potential energy of a charged particle in an electric field relies not only on the value of the electric field but also on the magnitude of the particle's charge.

Given information

The point charge is,\(Q = - 3.8\;{\rm{\mu C}}\).

The distance of the charge from the point b is,\({r_{\rm{b}}} = 88\;{\rm{cm}}\).

The distance of the charge from the point a is,\({r_{\rm{a}}} = 62\;{\rm{cm}}\).

Step 3:(a) Evaluation of the difference in the potential between point b and a

The electric potential at point b can be calculated as:

\(\begin{aligned}{V_{\rm{b}}} &= \frac{{kQ}}{{{r_{\rm{b}}}}}\\ &= \frac{{\left( {9 \times {{10}^9}\;{{{\rm{N}} \cdot {{\rm{m}}^{\rm{2}}}} \mathord{\left/{\vphantom {{{\rm{N}} \cdot {{\rm{m}}^{\rm{2}}}} {{{\rm{C}}^{\rm{2}}}}}} \right.} {{{\rm{C}}^{\rm{2}}}}}} \right)\left( {{\rm{ - 3}}{\rm{.8}}\;{\rm{\mu C}}} \right)\left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 6}}}}\;{\rm{C}}}}{{{\rm{1}}\;{\rm{\mu C}}}}} \right)}}{{\left( {{\rm{88}}\;{\rm{cm}}} \right)\left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 2}}}}\;{\rm{m}}}}{{{\rm{1}}\;{\rm{cm}}}}} \right)}}\\ &= - 3.9 \times {10^4}\;{\rm{V}}\end{aligned}\)

The electric potential at point a can be calculated as:

\(\begin{aligned}{V_{\rm{a}}} &= \frac{{kQ}}{{{r_{\rm{a}}}}}\\ &= \frac{{\left( {9 \times {{10}^9}\;{{{\rm{N}} \cdot {{\rm{m}}^{\rm{2}}}} \mathord{\left/{\vphantom {{{\rm{N}} \cdot {{\rm{m}}^{\rm{2}}}} {{{\rm{C}}^{\rm{2}}}}}} \right.} {{{\rm{C}}^{\rm{2}}}}}} \right)\left( {{\rm{ - 3}}{\rm{.8}}\;{\rm{\mu C}}} \right)\left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 6}}}}\;{\rm{C}}}}{{{\rm{1}}\;{\rm{\mu C}}}}} \right)}}{{\left( {62\;{\rm{cm}}} \right)\left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 2}}}}\;{\rm{m}}}}{{{\rm{1}}\;{\rm{cm}}}}} \right)}}\\ &= - 5.5 \times {10^4}\;{\rm{V}}\end{aligned}\)

The difference in the electric potential can be calculated as:

\(\begin{aligned}\Delta V &= {V_{\rm{b}}} - {V_{\rm{a}}}\\ &= \left( { - 3.9 \times {{10}^4}\;{\rm{V}}} \right) - \left( { - 5.5 \times {{10}^4}\;{\rm{V}}} \right)\\ &= 1.6 \times {10^4}\;{\rm{V}}\end{aligned}\)

Thus, the value of \({V_{\rm{b}}} - {V_{\rm{a}}}\)is \(1.6 \times {10^4}\;{\rm{V}}\).

Step 4:(b) Evaluation of the magnitude and direction of the net electric field

The electric field at point b can be calculated as:

\(\begin{aligned}{{{\bf{\vec E}}}_{\rm{b}}} &= \frac{{kq}}{{r_{\rm{b}}^{\rm{2}}}}\\ &= \frac{{\left( {9 \times {{10}^9}\;{{{\rm{N}} \cdot {{\rm{m}}^{\rm{2}}}} \mathord{\left/{\vphantom {{{\rm{N}} \cdot {{\rm{m}}^{\rm{2}}}} {{{\rm{C}}^{\rm{2}}}}}} \right.} {{{\rm{C}}^{\rm{2}}}}}} \right)\left( {{\rm{ - 3}}{\rm{.8}}\;{\rm{\mu C}}} \right)\left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 6}}}}\;{\rm{C}}}}{{{\rm{1}}\;{\rm{\mu C}}}}} \right)}}{{{{\left( {\left( {{\rm{88}}\;{\rm{cm}}} \right)\left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 2}}}}\;{\rm{m}}}}{{{\rm{1}}\;{\rm{cm}}}}} \right)} \right)}^2}}}\\ &= - 44.16 \times {10^3}\;{{\rm{V}} \mathord{\left/{\vphantom {{\rm{V}} {\rm{m}}}} \right.} {\rm{m}}}\end{aligned}\)

The electric field at point a can be calculated as:

\(\begin{aligned}{{{\bf{\vec E}}}_{\rm{a}}} &= \frac{{kq}}{{r_{\rm{a}}^{\rm{2}}}}\\ &= \frac{{\left( {9 \times {{10}^9}\;{{{\rm{N}} \cdot {{\rm{m}}^{\rm{2}}}} \mathord{\left/{\vphantom {{{\rm{N}} \cdot {{\rm{m}}^{\rm{2}}}} {{{\rm{C}}^{\rm{2}}}}}} \right.} {{{\rm{C}}^{\rm{2}}}}}} \right)\left( {{\rm{ - 3}}{\rm{.8}}\;{\rm{\mu C}}} \right)\left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 6}}}}\;{\rm{C}}}}{{{\rm{1}}\;{\rm{\mu C}}}}} \right)}}{{{{\left( {\left( {62\;{\rm{cm}}} \right)\left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 2}}}}\;{\rm{m}}}}{{{\rm{1}}\;{\rm{cm}}}}} \right)} \right)}^2}}}\\ &= - 88.96 \times {10^3}\;{{\rm{V}} \mathord{\left/{\vphantom {{\rm{V}} {\rm{m}}}} \right.} {\rm{m}}}\end{aligned}\)

The magnitude of the electric field can be calculated as:

\(\begin{aligned}\left| {{{{\bf{\vec E}}}_{\rm{b}}} - {{{\bf{\vec E}}}_{\rm{a}}}} \right| &= \sqrt {{\bf{\vec E}}_{\rm{b}}^{\rm{2}} + {\bf{\vec E}}_{\rm{a}}^2} \\ &= \sqrt {{{\left( { - 44.16 \times {{10}^3}\;{{\rm{V}} \mathord{\left/{\vphantom {{\rm{V}} {\rm{m}}}} \right.} {\rm{m}}}} \right)}^2} + {{\left( { - 88.96 \times {{10}^3}\;{{\rm{V}} \mathord{\left/{\vphantom {{\rm{V}} {\rm{m}}}} \right.} {\rm{m}}}} \right)}^2}} \\ &= 9.9 \times {10^4}\;{{\rm{V}} \mathord{\left/{\vphantom {{\rm{V}} {\rm{m}}}} \right.} {\rm{m}}}\end{aligned}\)

The direction of the electric field can be calculated as:

\(\begin{aligned}\theta &= {\tan ^{ - 1}}\left( {\frac{{{{{\bf{\vec E}}}_{\rm{a}}}}}{{{{{\bf{\vec E}}}_{\rm{b}}}}}} \right)\\ &= {\tan ^{ - 1}}\left( {\frac{{\left( { - 88.96 \times {{10}^3}\;{{\rm{V}} \mathord{\left/{\vphantom {{\rm{V}} {\rm{m}}}} \right.} {\rm{m}}}} \right)}}{{\left( { - 44.16 \times {{10}^3}\;{{\rm{V}} \mathord{\left/{\vphantom {{\rm{V}} {\rm{m}}}} \right.} {\rm{m}}}} \right)}}} \right)\\ &= 64^\circ \end{aligned}\)

Thus, the magnitude of the electric field is \(9.9 \times {10^4}\;{{\rm{V}} \mathord{\left/{\vphantom {{\rm{V}} {\rm{m}}}} \right.} {\rm{m}}}\) and direction is \(64^\circ \) above the positive x-axis.