Question

Point charges q1=+2.00μC and q2=−2.00μC are placed at adjacent corners of a square for which the length of each side is 4.50 cm . Point a is at the center of the square, and point b is at the empty corner closest to q2. Take the electric potential to be zero at a distance far from both charges.Part AWhat is the electric potential at point a due to q1 and q2?Part BWhat is the electric potential at point b?Part CA point charge q3 = -4.00 μC moves from point a to point b. How much work is done on q3 by the electric forces exerted by q1 and q2?

Answer 1

Answer:

A) 0 V

B) -117 kV

C) -0.468 J

Explanation:

q1=+2.00μC  q2=−2.00μC q3 = -4.00 μC

A) The electric potential at point a due to q1 and q2 ($V_a$) is given as:

$V_a=k\Sigma \frac{q}{r_i} = k(\frac{q_1}{r_1}+\frac{q_2}{r_2} )\\ but\ r_2=r_1=d.\ Therefore\\V_a=k(\frac{q_1}{d}+\frac{q_2}{d} )=k(\frac{2}{d}-\frac{2}{d} )=0$

B) he electric potential at point b due to q1 and q2 ($V_b$) is given as:

$V_b=k\Sigma \frac{q}{r_i} = k(\frac{q_1}{r_1}+\frac{q_2}{r_2} )\\ but\ r_2=4.5 cm=0.045m,\ r_1=\sqrt{0.045^2+0.045^2}= 0.0636.\ Therefore\\V_b= k(\frac{q_1}{r_1}+\frac{q_2}{r_2} )= 9*10^{9}(\frac{2*10^{-6}}{0.0636}-\frac{2*10^{-6}}{0.045} )=-117\ kV$

C) The work done on q3 by the electric forces exerted by q1 and q2 (W) is given by:

$W=q_3(V_a - V_b)=-4*10^{-6}(0-(-117*10^3))=-0.468\ J$