Name: Continuous Charge Distribution
Uploaded: 2019-07-19
Duration: 5 min 39 s
Description: This video will help us understand the necessity of continuous charge distribution and learn about the concept of linear charge density. Learn more.

Question

(i) State Coulomb's law and write the formula of electrostatic force between two charges, separated by certain distance. Under which condition this formula is applicable?
(ii) Two positive point charges which are

0.1 m

apart repel each other with a force of

18 N

. If the sum of the charges be

9 μ C

, calculate their separate values of charges.

Accepted Answer

(i) Coulomb's law is defined as, "The force of attraction or repulsion between two electric point charges is directly proportional to product of magnitude of their charges and inversely proportional to the square of the distance between the two charges"
The direction of force is along the line joining the point charges.
Consider two point charges $q_{1}$ and $q_{2}$ separated by distance $r$ in a medium of dielectric constant $k$ ,
If $F$ is the force between two point charges

$F \propto q_{1} q_{2} a n d F \propto 1 / r^{2}$

Therefore,

$F \propto \frac{q_{1} q_{2}}{r^{2}} F = \frac{k q_{1} q_{2}}{r^{2}}$

$F = \frac{1}{4 π \in_{\circ}} \frac{q_{1} q_{2}}{r^{2}}$

Where $\in_{\circ}$ is electrical permittivity of free space.

Here, $\frac{1}{4 π \in_{\circ}} = 9 \times 10^{9} N m^{2} C^{- 2}$

1. Coulomb's law holds good only for stationary charges.
2. It is applicable for point charges.

(ii) Let

$q_{1} = q μ C = q \times 10^{- 6} C$
$q_{2} = (9 - q) \times 10^{- 6} C$

As $F = \frac{q_{1} q_{2}}{4 π \in_{\circ} r^{2}}$

$∴ 18 = \frac{q \times 10^{- 6} \times (9 - q) \times 10^{- 6} \times 9 \times 10^{9}}{{(10^{- 1})}^{2}}$

$18 = (9 q - q^{2}) \frac{9}{10}$

$q^{2} - 9 q + 20 = 0, q^{2} - 5 q - 4 q + 20 = 0 q (q - 5) - 4 (q - 5) = 0 (q - 5) (q - 4) = 0 q = 5, 4$

So, the separate values of charges is $5 μ C & 4 μ C$

Important Questions on Electric Charges and Fields

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