Question

A spherical conductor of radius $10cm$ has a charge of $3.2\times {10}^{-7}C$ distributed uniformly. What is the magnitude of electric field at a point $15cm$ from the center of the sphere? $\left(\frac{1}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}=9\times {10}^{9}Nm/C^2\right)$

Answer 1

Step-1: Given data

1. The radius of the conductor is $R=0.1m.$
2. The charge in the spherical conductor is $Q=$$3.2\times {10}^{-7}C$.
3. The distance of the point where the electric field has to be find$15cm=0.15m.$

Step-2: Electric field

1. The electric field due to a conductor at a point R is defined by the form, $E=\frac{1}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}\frac{Q}{R^2}$, where, Q is the charge in the conductor and $\frac{1}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}$ is the Coulomb constant.

Step-3: Finding the electric field

As we know the electric field at the point R is $E=\frac{1}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}\frac{Q}{R^2}$.

So,

$$\begin{gathered} E=\frac{1}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}\frac{Q}{R^2}=\left(9\times {10}^9\right)\times \frac{3.2\times {10}^{-7}}{{\left(0.15\right)}^2} \\ orE=1.28\times {10}^{5}N/C. \end{gathered}$$

Therefore, the electric field at a point $15cm$ from the center of the sphere is $1.28\times {10}^{5}N/C.$