91Ó°ÊÓ

Mass is a fundamental quantity of the object which remains constant throughout the universe. The mass is related to the density of the material as,

\(m = \rho V\)

Here, \(\rho \) is the density of the material and \(V\) is the volume of the object.

The volume of the sphere is,

\(V = \frac{4}{3}\pi {r^3}\)

Here,\(r\)is the radius of the sphere.

From equation (1.1) and (1.2), the mass of the object is,

\(m = \frac{4}{3}\pi \rho {r^3}\)

The weight of the object is,

\({F_g} = mg\)

Using equation (1.3),

\({F_{_g}} = \frac{4}{3}\pi \rho {r^3}g\)

Substitute\(920{\rm{ kg}}/{{\rm{m}}^3}\)for\(\rho \),\(1.00{\rm{ }}\mu {\rm{m}}\)for\(r\), and\(9.81{\rm{ m}}/{{\rm{s}}^2}\)for g.

\(\begin{array}{c}{F_g} = \frac{4}{3}\pi \times \left( {920{\rm{ kg}}/{{\rm{m}}^3}} \right) \times {\left( {1.00{\rm{ }}\mu {\rm{m}}} \right)^3} \times \left( {9.81{\rm{ m}}/{{\rm{s}}^2}} \right)\\ = \frac{4}{3}\pi \times \left( {920{\rm{ kg}}/{{\rm{m}}^3}} \right) \times {\left[ {\left( {1.00{\rm{ }}\mu {\rm{m}}} \right) \times \left( {\frac{{{{10}^{ - 6}}{\rm{ m}}}}{{1{\rm{ }}\mu {\rm{m}}}}} \right)} \right]^3} \times \left( {9.81{\rm{ m}}/{{\rm{s}}^2}} \right)\\ = 3.78 \times {10^{ - 16}}{\rm{ N}}\end{array}\)

Hence, the weight of the drop is \(3.78 \times {10^{ - 16}}{\rm{ N}}\).

The weight of the oil drop is supported by the electrostatic force due to the electric field. Therefore,

\({F_g} = qE\)

Here,\(q\)is the magnitude of the charge on electron, and\(E\)is the electric field.

The expression for the electric field is,

\(E = \frac{{{F_g}}}{q}\)

Substitute\(3.78 \times {10^{ - 14}}{\rm{ N}}\) for\({F_g}\)and\(1.6 \times {10^{ - 19}}{\rm{ C}}\)for\(q\),

\(\begin{array}{c}E = \frac{{\left( {3.78 \times {{10}^{ - 14}}{\rm{ N}}} \right)}}{{\left( {1.6 \times {{10}^{ - 19}}{\rm{ C}}} \right)}}\\ = 2.36 \times {10^5}{\rm{ N}}/{\rm{C}}\end{array}\)

Hence, the strength of the electric field is \(2.36 \times {10^5}{\rm{ N}}/{\rm{C}}\).