Two point charges A and B of magnitude $+8 \times 10^{–6} C$ and $-8 \times 10^{–6} C$ respectively are placed at a distance d apart. The electric field at the middle point O between the charges is $6.4 \times 10^{4} NC^{–1}$. The distance ‘d’ between the point charges A and B is:

Question

A thin half ring of radius $35 \text{ cm}$ is uniformly charged with a total charge of $Q$ coulomb. If the magnitude of the electric field at centre of the half ring is $100 \text{ V/m}$, then the value of $Q$ is ______ $\text{nC}$.
($\epsilon_o = 8.85 \times 10^{-12} \text{ C}^2/\text{Nm}^2$ and $\pi = 3.14$)

Answer 1

To find the distance $d$ between the charges, we need to understand the behavior of the electric field created by point charges. When two charges are equal in magnitude but opposite in sign, like in this case, they form an electric dipole. The net electric field at any point on the perpendicular bisector of this dipole can be calculated using the formula:

The electric field $E$ due to a point charge $q$ at a distance $r$ is given by the equation:

E = \frac{k |q|}{r^2}

where $k$ is the Coulomb's constant, $k = 8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2$.

In this problem, there are two charges, +$8 \times 10^{-6}$ C and $-8 \times 10^{-6}$ C, placed at a distance $d$ apart. The electric field at point $O$, the midpoint between the charges, is given as $6.4 \times 10^4 \, \text{NC}^{-1}$.

Let's denote the distance from each charge to the point $O$ as $\frac{d}{2}$. At the midpoint, the electric fields due to each charge add up, as they are in the same direction (from positive to negative charge):

E_{\text{net}} = E_A + E_B = \frac{k \times 8 \times 10^{-6}}{(d/2)^2} + \frac{k \times 8 \times 10^{-6}}{(d/2)^2}

E_{\text{net}} = 2 \times \frac{k \times 8 \times 10^{-6}}{(d/2)^2}

The given electric field at point $O$ is $6.4 \times 10^4 \, \text{NC}^{-1}$, so:

2 \times \frac{8.99 \times 10^9 \times 8 \times 10^{-6}}{(d/2)^2} = 6.4 \times 10^4

Simplifying, we get:

\frac{8.99 \times 10^9 \times 8 \times 10^{-6}}{(d/2)^2} = 3.2 \times 10^4

Solving for $(d/2)^2$,

(d/2)^2 = \frac{8.99 \times 10^9 \times 8 \times 10^{-6}}{3.2 \times 10^4}

(d/2)^2 = \frac{7.192 \times 10^4}{3.2 \times 10^4}

(d/2)^2 = 2.2475 \times 10^{-2}

d/2 = \sqrt{2.2475 \times 10^{-2}}

d/2 = 0.15

d = 3 \times 0.15 = 0.3 \, \text{m}

Thus, the distance $d$ between the charges is 3.0 m.

Answer 2