Question

A small object of mass 4.18 g and charge -19 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?
magnitude

Answer 1

Concepts and reason

The concept of translational equilibrium and electric force are required to solve the problem.

First, derive the relation between the electric field and the weight of the object using the expression of electric force and the net force equations.

Then, substitute the values to find the magnitude and directions of the electric field.

Fundamentals

The electric force on a charged object can be given as,

$F = qE$

Here, q is the charge and E is the electric field.

An object is in translational equilibrium if the net force acting on the object is zero. That is,

$\Sigma F = 0$

The weight of an object is the force on an object due to gravity and it is given by the expression,

${F_{\rm{W}}} = mg$

Here, m is the mass of the object and g is acceleration due to gravity.

The weight of the object always points in the downward direction and is given by,

${F_{\rm{W}}} = mg$

The object is in equilibrium as it is floating in mid-air and is motion-less. This is possible only when an equal and opposite force is acting in upward direction. Thus, the electric force is acting in the upward-vertical direction.

The electric force is given by,

$F = qE$

Equate both forces and find E.

$\begin{array}{c}\\qE = mg\\\\E = \frac{{mg}}{q}\\\end{array}$

The expression of the electric field is as follows:

$E = \frac{{mg}}{q}$

Substitute 4.18 g for m, $9.8{\rm{ m/}}{{\rm{s}}^2}$ for g, and $19{\rm{ }}\mu {\rm{C}}$ for q in the above expression.

$\begin{array}{c}\\E = \frac{{\left( {4.18{\rm{ g}}} \right)\left( {9.8{\rm{ m/}}{{\rm{s}}^2}} \right)}}{{\left( {19{\rm{ }}\mu {\rm{C}}} \right)}}\left( {\frac{{1{\rm{ kg}}}}{{1000{\rm{ g}}}}} \right)\left( {\frac{{1{\rm{ }}\mu {\rm{C}}}}{{{{10}^{ - 6}}{\rm{ C}}}}} \right)\\\\ = 2.156 \times {10^3}{\rm{ N/C}}\\\end{array}$

The direction of electric force is in upward direction. The electric field is negative of the force divide by the charge magnitude. Thus, the direction of the electric field is opposite to the direction of the force.

Thus, the direction of the electric field is vertically downwards.

Ans:

The magnitude and the direction of the electric field are $2.156 \times {10^3}{\rm{ N/C}}$ and vertically-downwards respectively.