Chapter 16, Problem 70GP

Solution

(a)

To Find: Force exerted by sphere B on A.

  $F=\frac{k{Q}^2}{R^2}$

Given Information:

Both sphere A and B have same charge Q and are separated by a distance of R .

Formula Used:

By Coulomb’s law, force between two charges q and q’ which are r distance apart is:

  $F=\frac{kqq{\text{'}}^2}{r^2}$

Here, k is the Coulomb’s constant.

Calculation:

Both the spheres A and B have same charge. So, $q=q\text{'}=Q$ .

The distance between the charges is $r=R$ .

Hence, the force extend by B on A is

  $F_{AB}=\frac{k{Q}^2}{R^2}$

The force would be directed away from B as same charges repel each other.

(b)

To Find: Force on a due to B when B is bright in tow with another uncharged sphere C

  $\frac{k{Q}^2}{2R^2}$

Given Information:

Sphere B of charge Q is brought in touch with another uncharged sphere C. Sphere C is moved far away thereafter.

Formula Used:

By Coulomb’s law, force between two charges q and q’ which are r distance apart is:

  $F=\frac{kqq{\text{'}}^2}{r^2}$

Here, k is the Coulomb’s constant.

Calculation:

The sphere C is initially neutral.When it touches B, positive charges flow into C, till both the spheres have the same charge.The charges are shared between the two spheres.

Hence final charge on sphere B and C is $Q^{\prime }=\frac{Q}{2}$ .As sphereC is moved far away, the net force on A is due to B alone which is:

  $F_A\text{'}=\frac{kQ{Q}^1}{R^2}$

  $F_A\text{'}=\frac{k{Q}^2}{2R^2}$ away from B.

(c)

To Find: Force On due to B when sphere A is brought is touch with sphere C

  $\frac{3}{8}\frac{k{Q}^2}{R^2}$

Given Information:

Sphere C is made to touch sphere A and then it is moved far away.

Formula Used:

By Coulomb’s law, force between two charges q and q’ which are r distance apart is:

  $F=\frac{kqq{\text{'}}^2}{r^2}$

Here, k is the Coulomb’s constant.

Calculation:

When C makes contact with A, again charges flow from A to C,

Hence the final charge on A and C is $Q"=\frac{Q+\frac{Q}{2}}{2}$

Or, $Q"=\frac{3Q}{4}$

Now, as C is moved far away, The net force on A due to B is

  $\begin{array}{l}F"=\frac{kQ"Q\text{'}}{R^2}\hfill \\ F"=\frac{k(3Q/4)(Q/2)}{R^2}\hfill \end{array}$

  $F_2=\frac{3}{8}\frac{k{Q}^2}{R^2}$ away from B.

Conclusion:

Force on A due to B is $\frac{3}{8}\frac{k{Q}^2}{R^2}$ and away from B.