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• Notes

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 $E ?$

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

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

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

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

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

$=-4.8 * 10^{-5}= J$