91影视

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.

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,\({r_{\rm{a}}} = 62\;{\rm{cm}}\).

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 \(1.6 \times {10^4}\;{\rm{V}}\).

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.