Tag: exchange of electrons

Questions Related to exchange of electrons

The linear charge density of a thin metallic rod varies with the distance $'x'$ from one end as $\lambda  = {\lambda _0}{x^2}\left( {0 \leqslant x \leqslant l} \right).$ The total charge on the rod is:

physics static electricity exchange of electrons methods used for charging electric charge

  1. $\dfrac{{{\lambda _0}{l^3}}}{3}$

  2. $\dfrac{{{\lambda _0}{l^4}}}{3}$

  3. $\dfrac{{2{\lambda _0}{l^3}}}{3}$

  4. $\dfrac{{{\lambda _0}{l}}}{2}$

Show answer
Correct Option: D
Explanation:
$\lambda =\dfrac {\lambda _0 x}{L}$
$Q=\displaystyle \int _0^L \lambda \ dx$
$=\displaystyle \int _0^L \dfrac {\lambda _0 x}{L}dx$
$=\dfrac {\lambda _0}{L} \dfrac {x^2}{2}\displaystyle \int _0^L$
$=\dfrac {\lambda _0}{L}\times \dfrac {L^2}{2}$
$=\dfrac {\lambda _0L}{2}$

A hollow metal sphere, Sphere A, sits on an insulating stand. Sphere A has a diameter of 4 inches, and a net charge of magnitude $Q _0$. A second hollow metal sphere, Sphere B, also sits on an insulating stand, but has a diameter of 8 inches and zero net charge. The two spheres are brought close so that they touch, then they are separated.
In terms of $Q _0$, what is the final charge on Sphere A?

physics static electricity exchange of electrons methods used for charging electric charge

  1. $\cfrac{Q _0}{5}$

  2. $\cfrac{Q _0}{4}$

  3. $\cfrac{Q _0}{2}$

  4. $Q _0$

  5. $4Q _0$

Show answer
Correct Option: A
Explanation:

$R _B=2R _A$


Let final charge on spheres be $Q _A$ and $Q _B$.

On touching the spheres charge density becomes same.

And, charge density, $\sigma=\dfrac{Q}{4\pi R^2}$

$\implies Q\propto R^2$ for same $\sigma$

Hence, $\dfrac{Q _B}{Q _A}=\dfrac{R _B^2}{R _A^2}=2^2$

$\implies Q _B=4Q _A$

And, total charge$=Q _B+Q _A=Q _o$

$\implies 4Q _A+Q _A=Q _o$

$\implies Q _A=\dfrac{Q _o}{5}$

Answer-(A)