Answer to Question #210189 in Electric Circuits for kilo

Question

Electric Potential and Electric Potential Energy. 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 the figure below. (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 q3= 3.00 μC as the latter charge moves from infinity to point P.

Answer 1

Answer:-

$q_1=2\times 10^{-6} \ C\\ q_2=-6\times10^{-6} \ C\\ q_3=3\times 10^-{-6} \ C$

(1)

$r_1=4m , \ \ \ \ \ r_2 =\sqrt{3^2+4^2}=5m\\ V_p=\frac{kq_1}{r_1}+\frac{kq_2}{r_2}$

$V_p=\frac{(8.99\times10^9)\times (2.00\times10^{-6})}{4}+\frac{(8.99\times 10^9)\times (-6\times 10^{-6})}{5}$

$\boxed{V_p=-6293 \ volt}$

(2) At infinity i.e $V_{\infin}=0$

hence $\Delta U =q(\Delta V)\\$

$\implies \Delta U=q(V_p-V_{\infin})$

$\implies \Delta U=(3\times10^{-6})(-6293-0)$

$\implies \boxed{\Delta U=-0.0189 \ joule}$