Electrostatics Exam 1 and Problem Solutions

1. If we touch two spheres to each other, find the final charges of the spheres. Electrostatics 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. Electrostatics A and C are negatively charged by induction. Thus, leaves of both electroscopes are negatively charged. Electrostatics 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. Electrostatics Free body diagram of forces is given below; Electrostatics 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. Electrostatics We assume that there is a +q charge at point A while finding electric field at point A. Electrostatics 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. Electrostatics 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