find the Potential energy at one of the charges due to the distance and othter charge.
PE=kQ1Q2/d
What is the PE at 4 cm closer?
PE=kQ1Q2/(d-.04)
then thw work moving Q2 must be
difference in PE
two positive points charges of 15c and 13c are placed 12cm apart . find the work done in bringing the two charges 4cm closer.
2 answers
W=u¹+u²
u¹=kq¹q²/d
u²=kq¹q²/d-0.5
u¹=kq¹q²/d
u²=kq¹q²/d-0.5
0 answers