Question

A metal sphere of radius 10 cm carries a charge of +2.0 µC. What is the magnitude of the electric field 5.0 cm outside the sphere's surface?

Answer 1

Concepts and reason

The concept used to solve this question is the electric field due to a uniformly charged sphere.

Initially, explain the mathematical formula for finding the electric field due to a uniformly charged sphere. After that, using the formula of the electric field due to a uniformly charged sphere determine the magnitude of the electric field at the given point.

Fundamentals

Electric field due to uniformly charged sphere:

The electric field due to the charged sphere can be expressed as,

$E = \frac{{kQ}}{{{r^2}}}$

Here, $E$ is the electric field due to the charged sphere, $Q$ is the magnitude of charge carried by the sphere, $r$ is the distance from the center of the sphere and the point at which the field is calculated and $k$ is a constant

The electric field due to the charged sphere can be expressed as,

$E = \frac{{kQ}}{{{r^2}}}$

Here, $E$ is the electric field due to the charged sphere, $Q$ is the magnitude of charge carried by the sphere, $r$ is the distance from the center of the sphere and the point at which the field is calculated and $k$ is a constant.

Value of constant $k$ is $\frac{1}{{4\pi {\varepsilon _0}}}$ .

Mathematically, the electric field at a point outside the sphere can be calculated using,

$E = \frac{{kQ}}{{{r^2}}}$

The point is located at a point $5{\rm{ cm}}$ outside the sphere.

So the value of $r$ can be evaluated as,

$\begin{array}{l}\\r = 10{\rm{ cm + 0}}{\rm{.5 cm}}\\\\r = 15{\rm{ cm}}\left( {\frac{{1{\rm{ m}}}}{{{{10}^{2.}}{\rm{ cm}}}}} \right)\\\\r = 0.15{\rm{ m}}\\\end{array}$

Substitute $9 \times {10^9}{\rm{N}} \cdot {{\rm{m}}^2} \cdot {{\rm{C}}^{ - 2}}$ for $k$ , $+ 2.0{\rm{ }}\mu {\rm{C}}$ for $Q$ and $0.15{\rm{ cm}}$ for $r$

$\begin{array}{l}\\E = \frac{{9 \times {{10}^9}{\rm{N}} \cdot {{\rm{m}}^2} \cdot {{\rm{C}}^{ - 2}}\left( { + 2.0{\rm{ }}\mu {\rm{C}}} \right)\left( {\frac{{1\,{\rm{C}}}}{{{{10}^6}\mu C}}} \right)}}{{{{\left( {0.15{\rm{ m}}} \right)}^2}}}\\\\E = 800{\rm{ kN/C}}\\\end{array}$

The value of the electric field is calculated as $800{\rm{ kN/C}}$ .

Ans:

The value of the electric field outside the point is $800{\rm{ kN/C}}$ .