**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;**

**q _{r}=(Q_{1}+Q_{2})/(r_{1}+r_{2})**

**q _{r}=(20-5)q/(2r+r)=5q/r**

**Charge of first sphere becomes;**

**Q _{1}=q_{r}.r_{1}=5q/r.2r=10q**

**Charge of second sphere becomes;**

**Q _{2}=q_{r}.r_{2}=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/d ^{2}=k.Q^{2}/d^{2}**

**F _{C}=k.Q.(-4Q)/(√2d)^{2}=-4k.Q^{2}/2d^{2}=-2.k.Q^{2}/d^{2}=-2F**

**F _{D}=k.Q.2Q/d^{2}=2k.Q^{2}/d^{2}=2F**

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**4.** Find the electric field at point A produced by charges q_{1} and q_{2} 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.

**E _{1}=k.(-4q)/d^{2}=-4k.q/d^{2}**

**E _{2}=k.(16q)/4d^{2}=4k.q/d^{2}**

**Resultant electric field at point A is;**

**E _{resultant}=E_{1}+E_{2}=-4kq/d^{2}+4kq/d^{2}=0**

**5.** Find electric potential energy produced by Q_{1}, Q_{2} and Q_{3} in terms of k.q^{2}/r.

**Ep=k.Q _{1}.Q_{2}/r**

**E _{P1,2}=k.10q.(-12q)/3r=-40k.q^{2}/r**

**E _{P1,3}=k.10q.5q/4r=25k.q^{2}/2r**

**E _{P2,3}=k.(-12q).5q/5r=-12k.q^{2}/r**

**E _{P}=E_{P1,2+}E_{P1,3+}E_{P2,3}**

**E _{P}=-40k.q^{2}/r+25k.q^{2}/2r-12k.q^{2}/r**

**E _{P}=-39,5k.q^{2}/r**

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