Name: Dielectrics in a Capacitor
Uploaded: 2019-07-19
Duration: 6 min 3 s
Description: Let us use our knowledge about dielectrics and capacitors to find out how the use of dielectrics can increase the capacitance of a system. Learn through this...

Question

Two positive charges,

q_{1}

and

q_{2}

, are located at the points with radius vectors

\vec{r_{1}}

and

\vec{r_{2}}

, respectively. Find a negative charge

q_{3}

and a radius vector

\vec{r_{3}}

of the point at which it has to be placed for the force acting on each of the three charges to be equal to zero.

Accepted Answer

Let the three point charges, $q_{1}, q_{2}$ and $q_{3}$ , having position vectors, $\vec{r_{1}}, \vec{r_{2}}$ and $\vec{r_{3}}$ , respectively, be fixed in the coordinate system, as shown in the figure.

For equilibrium, the net coulomb force on each charge must be zero.

For charge $q_{3}$ ,

${\vec{F}}_{32} + {\vec{F}}_{31} = 0$

$\frac{+ q_{2} q_{3} (\vec{r_{2}} - \vec{r_{3}})}{{|\vec{r_{2}} - \vec{r_{3}}|}^{3}} + \frac{q_{1} q_{3} (\vec{r_{1}} - \vec{r_{3}})}{{|\vec{r_{1}} - \vec{r_{3}}|}^{3}} = 0$

$\Rightarrow \frac{q_{2}}{{|\vec{r_{2}} - \vec{r_{3}}|}^{2}} = \frac{q_{1}}{{|\vec{r_{1}} - \vec{r_{3}}|}^{2}} . . . (1)$

As we can see in the figure,

$\frac{\vec{r_{2}} - \vec{r_{3}}}{|\vec{r_{2}} - \vec{r_{3}}|} = - \frac{\vec{r_{1}} - \vec{r_{3}}}{|\vec{r_{1}} - \vec{r_{3}}|}$

From equation $(1)$ ,

$\sqrt{q_{2}} (\vec{r_{1}} - \vec{r_{3}}) = \sqrt{q_{1}} (\vec{r_{3}} - \vec{r_{2}})$

$\Rightarrow \vec{r_{3}} = \frac{\sqrt{q_{2}} \vec{r_{1}} + \sqrt{q_{1}} \vec{r_{2}}}{\sqrt{q_{1}} + \sqrt{q_{2}}} . . . (2)$

Similarly, for the equilibrium of charge $q_{1},$

${\vec{F}}_{13} + {\vec{F}}_{12} = 0$

$\Rightarrow \frac{q_{3} (\vec{r_{3}} - \vec{r_{1}})}{{|\vec{r_{3}} - \vec{r_{1}}|}^{3}} + \frac{q_{2} (\vec{r_{2}} - \vec{r_{1}})}{{|\vec{r_{2}} - \vec{r_{1}}|}^{3}} = 0$

$\Rightarrow q_{3} = \frac{- q_{2}}{{|\vec{r_{2}} - \vec{r_{1}}|}^{2}} {|\vec{r_{1}} - \vec{r_{3}}|}^{2}$

Substituting the value of $\vec{r_{3}}$ from equation $(2)$ , we get,

$q_{3} = \frac{- q_{1} q_{2}}{{(\sqrt{q_{1}} + \sqrt{q_{2}})}^{2}}$

Important Questions on ELECTRODYNAMICS

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Two positive charges, q1 and q2, are located at the points with radius vectors r1→ and r2→, respectively. Find a negative charge q3 and a radius vector r3→ of the point at which it has to be placed for the force acting on each of the three charges to be equal to zero.

Important Questions on ELECTRODYNAMICS

Learn and Prepare for any exam you want !