Question

A\(5.00{\rm{ g}}\)charged insulating ball hangs on a\(30.0{\rm{ cm}}\)long string in a uniform horizontal electric field as shown in Figure 18.56. Given the charge on the ball is\(1.00{\rm{ }}\mu {\rm{C}}\), find the strength of the field.

Figure 18.56 A horizontal electric field causes the charged ball to hang at an angle of\(8.00^\circ \).

Short answer

The strength of the electric field is \(6.89 \times {10^3}{\rm{ N}}/{\rm{C}}\).

Step-by-step solution

Electrostatic force

When a charged particle is placed in an electric, it experiences some electrostatic force.The expression for the force experienced by the charged particle is,

\({F_e} = qE\)

Here, \(q\) is the charge on the proton and \(E\) is the electric field.

Free body diagram of charged insulating ball

The free body diagram or the force diagram of the charged insulating ball when it is placed in an electric field is represented as,

Free body diagram of the charged insulating ball when it is placed in an electric field

Here, \(q\) is the charge on the insulating ball, \(E\) is the electric field \(qE\) is the force experienced by the charged insulating ball, \(m\) is the mass of the ball, \(g\) is the acceleration due to gravity, \(mg\) is the weight of the ball, and \(T\) is the tension force in the string.

Force equations

The equilibrium force equation in the horizontal direction is,

\(\begin{array}{c}\sum {{F_x}} = 0\\qE - T\sin \left( {{8^ \circ }} \right) = 0\\T\sin \left( {{8^ \circ }} \right) = qE\end{array}\)

The equilibrium force equation in the vertical direction is,

\(\begin{array}{c}\sum {{F_y}} = 0\\T\cos \left( {8^\circ } \right) - mg = 0\\T\cos \left( {8^\circ } \right) = mg\end{array}\)

Dividing equation (1.1) by (1.2),

\(\begin{array}{c}\frac{{T\sin \left( {8^\circ } \right)}}{{T\cos \left( {8^\circ } \right)}} = \frac{{qE}}{{mg}}\\\tan \left( {8^\circ } \right) = \frac{{qE}}{{mg}}\end{array}\)

Rearranging the above equation in order to get an expression for the electric field.

\(E = \frac{{mg\tan \left( {8^\circ } \right)}}{q}\)

Substitute\(5.00{\rm{ g}}\)for\(m\),\(9.81{\rm{ m}}/{{\rm{s}}^2}\)for\(g\)and\(1.00{\rm{ }}\mu {\rm{C}}\)for\(q\).

\(\begin{array}{c}E = \frac{{\left( {5.00{\rm{ g}}} \right) \times \left( {9.81{\rm{ m}}/{{\rm{s}}^2}} \right) \times \tan \left( {8^\circ } \right)}}{{\left( {1.00{\rm{ }}\mu {\rm{C}}} \right)}}\\ = \frac{{\left( {5.00{\rm{ g}}} \right) \times \left( {\frac{{{{10}^{ - 3}}{\rm{ kg}}}}{{1{\rm{ g}}}}} \right) \times \left( {9.81{\rm{ m}}/{{\rm{s}}^2}} \right) \times \tan \left( {8^\circ } \right)}}{{\left( {1.00{\rm{ }}\mu {\rm{C}}} \right) \times \left( {\frac{{{{10}^{ - 6}}{\rm{ C}}}}{{1{\rm{ }}\mu {\rm{C}}}}} \right)}}\\ = 6.89 \times {10^3}{\rm{ N}}/{\rm{C}}\end{array}\)

Hence, the strength of the electric field is \(6.89 \times {10^3}{\rm{ N}}/{\rm{C}}\).