A charge q1 = 2.00 μC is located at the origin, and a charge q2 = -6.00 μC is located at (0, 3.00) m, as shown in Figure 25.12a. (A) Find the total electric potential due to these charges at the point P, whose coordinates are (4.00, 0) m. (B) Find the change in potential energy of the system of

Question

A charge [latex] q_{1}=2.00 \mu C [/latex] is located at the origin, and a charge [latex] q_{2}=-6.00 \mu C [/latex] is located at (0, 3.00) m, as shown in Figure 25.12a.

(A) Find the total electric potential due to these charges at the point P, whose coordinates are (4.00, 0) m.

(B) Find the change in potential energy of the system of two charges plus a charge [latex] q_{3}=3.00 \mu C [/latex] as the latter charge moves from infinity to point P (Fig. 25.12b).

Accepted Answer

For two charges, the sum in Equation 25.12 gives

[latex] V=k_{e} \sum_{i} \frac{q_{i}}{r_{i}} [/latex] (25.12).

[latex] V_{P}=k_{e}\left(\frac{q_{1}}{r_{1}}+\frac{q_{2}}{r_{2}} ight) [/latex].

[latex] V_{P}=\left(8.99 imes 10^{9} N \cdot m ^{2} / C ^{2} ight) [/latex]

[latex] imes\left(\frac{2.00 imes 10^{-6} C }{4.00 m }-\frac{6.00 imes 10^{-6} C }{5.00 m } ight) [/latex].

[latex] =-6.29 imes 10^{3} V [/latex].

(B) Find the change in potential energy of the system of two charges plus a charge [latex] q_{3}=3.00 \mu C [/latex] as the latter charge moves from infinity to point P (Fig. 25.12b).

[latex] ext { When the charge } q_{3} ext { is at infinity, let us define } [/latex][latex] U_{i}=0 ext { for the system, and when the charge is at } P [/latex], [latex] U_{f}=q_{3} V_{P} ; ext { therefore } [/latex]

[latex] \Delta U=q_{3} V_{P}-0=\left(3.00 imes 10^{-6} C ight)\left(-6.29 imes 10^{3} V ight) [/latex].

[latex] =-1.89 imes 10^{-2} J [/latex].

Therefore, because the potential energy of the system has decreased, positive work would have to be done by an external agent to remove the charge from point P back to infinity.

Question 25.3: A charge q1 = 2.00 μC is located at the origin, and a charge...

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