Define the electric field strength at a point in an electric field - AQA - A-Level Physics - Question 2 - 2018 - Paper 2

In this scenario, the electric field strengths produced by both charges at point P must be equal and opposite since the resultant electric field strength is zero. The electric field due to a point charge is given by:

$E = \frac{k \cdot |Q|}{r^2}$

Where:

• $E$ is the electric field strength
• $k$ is Coulomb's constant (approximately $8.99 \times 10^9 \text{ Nm}^2/\text{C}^2$)
• $|Q|$ is the magnitude of the charge that produces the electric field
• $r$ is the distance from the charge to the point.

Let $E_Q$ be the electric field due to charge Q and $E_{46}$ be the electric field due to +46 μC. At position P:

$E_Q = E_{46}$

Hence,

$\frac{k \cdot |Q|}{66^2 \times 10^{-6}} = \frac{k \cdot 46 \times 10^{-6}}{120^2 \times 10^{-6}}$

By canceling $k$ and rearranging, we find:

$|Q| = 46 \times \frac{66^2}{120^2} = 46 \times \frac{4356}{14400} \approx 13.93 \, \mu C.$

Work must be done on the positive proton because it is at a positive potential when compared to infinity. Since the proton is moving against the repulsive forces from the like charges, work will be required to move it to position P.

When the thread breaks, the rubber ball will experience an unbalanced force due to the horizontal electric field. Initially, it is in equilibrium, balancing the tensions from the thread. However, after the thread breaks, the only force acting on the ball is the electric force in the horizontal direction, causing it to accelerate horizontally away from the wall.