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A uniform electric field has magnitude \(E\) and is directed in
the negative \(x\) -direction. The potential difference between point \(a\)
(at \(x=0.60\) m) and point \(b(\text { at } x=0.90 \mathrm{m})\) is 240 V . (a) Which
point, \(a\) or \(b\) , is at the higher potential? (b) Calculate the value of \(E\) .
(c) A negative point charge \(q=-0.200 \mu \mathrm{C}\) is moved from \(b\) to \(a\) .
Calculate the work done on the point charge by the electric field.

*\( V=\frac{u}{q_0} \)*

*\( \frac{W_{a \rightarrow b}}{q_{0}}=V_{a}-V_{b} \)*

*\( V=\frac{q}{4 \pi \varepsilon_{r}} \)*

*\( V_{a}-V_{b}=\int_{a}^{b} E \cdot d l \)*

A uniform electric field has magnitude \(E\) and is directed in
the negative \(x\) -direction. The potential difference between point \(a\)
(at \(x=0.60\) m) and point \(b(\text { at } x=0.90 \mathrm{m})\) is 240 V . (a) Which
point, \(a\) or \(b\) , is at the higher potential? (b) Calculate the value of \(E\) .
(c) A negative point charge \(q=-0.200 \mu \mathrm{C}\) is moved from \(b\) to \(a\) .
Calculate the work done on the point charge by the electric field.\( K_{A}+U_{A}=K_{B}+U_{B} \)

\( U=q V \longrightarrow K_{A}+q V_{A}=K_{B}+q V_{B} \)

\( K_{B}=K_{A}+q\left(V_{A}-V_{B}\right) =\frac{1}{2} m K_{A}^{2}+q\left(V_ A-V_{B}\right) \)

\( =0.0025+\left(-5 * 10^{-6}\right)(200-800)=0.0055 \mathrm{J} \)

\(∴ K_{B}=0.0055 \)

\( K_{B}=\frac{1}{2} m V_{B}^{2} \longrightarrow V_{B} =\sqrt{\frac{2 K_{B}}{m}} =\sqrt{\frac{2 * 0.0055}{2*10^{-4}}} =7.42 \mathrm{mi} \)

A particle with a charge of \(+4.20 \mathrm{nC}\) is in a uniform elec-
tric field \(\vec{E}\) directed to the left. It is released from rest and moves to
the left; after it has moved 6.00 cm, its kinetic energy is found to
be \(+1.50 \times 10^{-6} \mathrm{J}\) . (a) What work was done by the electric force?
(b) What is the potential of the starting point with respect to the
end point? (c) What is the magnitude of \(E ?\)

\( W_{a \rightarrow b}=F_{s} \cos \theta=E|q| s \cos \theta \)

(a) \( W_{a \rightarrow b}=K_{b}-K_{0}=1.5 * 10^{-6}-0=1.5*10^{-6} \mathrm{J} \)

(b) \( V_{a}-V_{b}=\frac{W_{a\rightarrow b}}{q}= \frac{1.5 * 10^{-6}}{4 \cdot 2 * 10^{-9}}=357 \mathrm{V} \)

\( W_{a \rightarrow b}=E|q| s \rightarrow E=\frac{W_{a \rightarrow b}}{|q|s}=\frac{1.5*10_6}{4.2*10^9*0.06} \)

\( 5.95*10^{3} \mathrm{v/m} \)

OR \( E=\frac{V_{a}-V_{b}}{s} =\frac{357}{0.06}=5.95*10^{3} \mathrm{v} / \mathrm{m} \)

A particle with a charge of \(+4.20 \mathrm{nC}\) is in a uniform elec-
tric field \(\vec{E}\) directed to the left. It is released from rest and moves to
the left; after it has moved 6.00 cm, its kinetic energy is found to
be \(+1.50 \times 10^{-6} \mathrm{J}\) . (a) What work was done by the electric force?
(b) What is the potential of the starting point with respect to the
end point? (c) What is the magnitude of \(E ?\)

\( V_{a-b}=\int_{a}^{b} \vec{E} \cdot d \vec{l} \)

\( V_{b}-V_{a}= -\int_{a}^{b} \vec{E} \cdot \overrightarrow{d L} =\int_{a}^{b} E \cdot d x=E\left(x_{b}-x_{n}\right) \)

\( V_{b} -V_{a} =E\left(X_{b} \cdot X_{a}\right) \rightarrow E=\frac{V_{b}-V_{a}}{X_{b}-X_{a}} \)

\(∴ E=\frac{240}{0.9-0.6}=800 \mathrm{v/m} \)

\( W_{b \rightarrow a}=q\left(V_{b}-V_{a}\right)=\left(-0.2 * 10^{-6}\right)\left(240\right) \)

\( =-4.8 * 10^{-5}= J\)

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