Question

A point charge q2 = -3.8 μC is fixed at the origin of a co-ordinate system as shown. Another point charge q1 = 1.7 μC is is initially located at point P, a distance d1 = 8.5 cm from the origin along the x-axis

The charge q2 is now replaced by two charges q3 and q4 which each have a magnitude of -1.9 μC, half of that of q2. The charges are located a distance a = 2 cm from the oringin along the y-axis as shown. What is ΔPE, the change in potential energy now if charge q1 is moved from point P to point R?

Answer 1

Answer:

The potential energy is

ΔPE=15.39 N

Explanation:

In the initial sequence of the charge the energy initial is

$PE=k*\frac{abs(q1*q2)}{r^{2}} \\PE=8.85x10^{9} *\frac{abs(-3.8uC*1.7uC)}{(8.5cm)^{2}} \\PE=8.85x10^{9}\frac{N*m^{2} }{C^{2} }*\frac{3.8uC*1.7uC}{(0.085m)^{2}} \\\\PE=8.85x10^{9}\frac{N*m^{2} }{C^{2} } *\frac{6.46uC^{2} }{7.225x10^{-3} m^{2}} \\\\PE=7.9N$

The change the charge and the distance the second energy is

r=$\sqrt{2^{2} +8.5^{2} }=8.732cm\\ 8.732cm\frac{1m}{100cm} =87.32x10^{-3}m$

$PE=k*\frac{abs(q1*q3*q4)}{r^{2}} \\PE=8.85x10^{9} *\frac{abs(-1.9uC*-1.9uC*1.7uC)}{(8.732cm)^{2}} \\PE=8.85x10^{9}\frac{N*m^{2} }{C^{2} }*\frac{1.9uC*1.9uC*1.7uC}{(0.08732m)^{2}} \\\\PE=8.85x10^{9}\frac{N*m^{2} }{C^{2} } *\frac{6.46uC^{2} }{7.624x10^{-3} m^{2}} \\\\PE=7.49N$

And finally the total energy is adding both so

ΔPE=7.49N+7.9N

ΔPE=15.39 N