Question

Calculate the electric field strength required to just support a water drop of mass ${10^{ - 7}}kg$ and having a charge $1.6 \times {10^{ - 19}}C$.

Answer 1

The weight of the water drop will be acting normally downwards, causing the drop to fall. If we want the drop to be supported, an equal magnitude of force must be applied opposite to the weight of the drop. This equal and opposite force will be provided by the electric field (since our drop also carries a charge). The step by step solution of this methodology is given below.

Complete step-by-step solution:
Weight of the drop $(W) = mg = {10^{ - 7}}kg \times 9.8m/{s^2} = 9.8 \times {10^{ - 7}}N$
Let the magnitude/strength of the electric field be $E$
Force exerted on the drop by the electric field present in the region $(F) = qE \Rightarrow F = 1.6 \times {10^{ - 19}}C \times E$
To support the drop, the forces must balance each other out

$$\Rightarrow W = F \\\ \Rightarrow 9.8 \times {10^{ - 7}}N = 1.6 \times {10^{ - 19}}C \times E \\\ \Rightarrow E = \dfrac{{9.8 \times {{10}^{ - 7}}N}}{{1.6 \times {{10}^{ - 19}}C}} = 6.125 \times {10^{12}}N/C \\\$$

Hence, the electric field strength should be $6.125 \times {10^{12}}N/C$

Additional Information:
An electric field exerts a force on charges. If the charge is positive, the force is exerted against the electric field/away from the field whereas if the charge is negative, the force is exerted towards the field.
Two charges are always needed to visualise or encounter a force. In addition to the source charge, the charge that is used to measure the field strength is known as the test charge. Recalling that electric field is the force acting per unit charge, this force is exerted by the given charge (source charge) on the test charge.

Note:- For a body to be in equilibrium, the net force acting on the body must be zero. Algebraically, $\sum F = 0$ $\Rightarrow F + ( - W) = 0 \Rightarrow F = W$
This is how we can alternatively proceed with our solution.