Question

A small object of mass 3.96 g and charge -17.1 µC is suspended motionless above the ground when immersed in a uniform electric field perpendicular to the ground. What are the magnitude and direction of the electric field?
What is the magnitude _________ N/C
What is the direction (upward East West downward)

Answer 1

Concepts and reason

The main concepts used to solve the problem are force due to electric field and weight force.

Initially, use the expression of force acting on the charge in electric field. Later, find the expression of force acting on the charge due to its mass. Finally, equate the both force by using the fact that to keep the object in equilibrium both forces must be equal.

Fundamentals

The force acting on a charge in a uniform electric field is,

$F = qE$

Here, q is the charge, E is the electric field, and $F$ is the force acting due to electric field.

The force due to mass of the objects is,

$F = mg$

Here, m is the mass, g is the gravity, and F is the force due to mass.

The force acting on a charge in a uniform electric field is,

$F = \left| q \right|E$

Here, $\left| q \right|$ is the charge, E is the electric field, and $F$ is the force acting due to electric field.

The force due to mass of the objects is,

$F = mg$

Here, m is the mass, g is the gravity, and F is the force due to mass.

Equate the forces due to electric field and weight and find the expression of electric field.

$\begin{array}{c}\\\left| q \right|E = mg\\\\E = \frac{{mg}}{{\left| q \right|}}\\\end{array}$

Convert the mass from gram to kilogram by dividing it with 1000.

$\begin{array}{c}\\m = \left( {3.96\,{\rm{g}}} \right)\left( {\frac{{1\,{\rm{kg}}}}{{1000\,{\rm{g}}}}} \right)\\\\ = 3.96 \times {10^{ - 3}}\,{\rm{kg}}\\\end{array}$

Convert the charge from micro coulomb’s to coulomb’s by multiplying it with ${10^{ - 6}}$ .

$\begin{array}{c}\\q = \left( { - 17.1\,{\rm{\mu C}}} \right)\left( {\frac{{{{10}^{ - 6}}\,{\rm{\mu C}}}}{{1\,{\rm{\mu C}}}}} \right)\\\\ = - 17.1 \times {10^{ - 6}}\,{\rm{C}}\\\end{array}$

Substitute $3.96 \times {10^{ - 3}}\,{\rm{kg}}$ for m , $- 17.1 \times {10^{ - 6}}\,{\rm{C}}$ for q , and $9.8\,{\rm{m/}}{{\rm{s}}^{\rm{2}}}$ for g in expression $E = \frac{{mg}}{{\left| q \right|}}$ .

$\begin{array}{c}\\E = \frac{{\left( {3.96 \times {{10}^{ - 3}}\,{\rm{kg}}} \right)\left( {9.8\,{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right)}}{{\left| { - 17.1 \times {{10}^{ - 6}}\,{\rm{C}}} \right|}}\\\\ = 2269.5\,{\rm{N/C}}\\\end{array}$

Ans:

The magnitude and direction of the electric field are $2269.5\,{\rm{N/C}}$ and downward.