Electrostatics Exam1 and Problem Solutions
1. If we touch two spheres to each other, find the final charges of the spheres.
Charge per unit radius is found;
qr=(Q1+Q2)/(r1+r2)
qr=(20-5)q/(2r+r)=5q/r
Charge of first sphere becomes;
Q1=qr.r1=5q/r.2r=10q
Charge of second sphere becomes;
Q2=qr.r2=5q/r.r=5q
2. Positively charged sphere B is placed between two neutral spheres A and C. We cut connection of A and C with ground. If we put A closer to the first electroscope and touch C to the sphere of second electroscope, find the type of charge electroscopes have.
A and C are negatively charged by induction. Thus, leaves of both electroscopes are negatively charged.
3. If force applied by charge placed at point B on A is F, find forces applied by charges C and D on A in terms of F.
Free body diagram of forces is given below;
F=k.Q.Q/d2=k.Q2/d2
FC=k.Q.(-4Q)/(√2d)2=-4k.Q2/2d2=-2.k.Q2/d2=-2F
FD=k.Q.2Q/d2=2k.Q2/d2=2F
4. Find the electric field at point A produced by charges q1 and q2 in terms of k, q and d.
We assume that there is a +q charge at point A while finding electric field at point A.
E1=k.(-4q)/d2=-4k.q/d2
E2=k.(16q)/4d2=4k.q/d2
Resultant electric field at point A is;
Eresultant=E1+E2=-4kq/d2+4kq/d2=0
5. Find electric potential energy produced by Q1, Q2 and Q3 in terms of k.q2/r.
Ep=k.Q1.Q2/r
EP1,2=k.10q.(-12q)/3r=-40k.q2/r
EP1,3=k.10q.5q/4r=25k.q2/2r
EP2,3=k.(-12q).5q/5r=-12k.q2/r
EP=EP1,2+EP1,3+EP2,3
EP=-40k.q2/r+25k.q2/2r-12k.q2/r
EP=-39,5k.q2/r
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