Two small identical spheres, A and B, each carrying a charge of +5 µC, are placed 2 m apart - NSC Physical Sciences - Question 8 - 2017 - Paper 1

In the diagram, the electric field lines emanate outward from both spheres A and B since both are positively charged. Each field line represents the direction of the force that a positive test charge would experience. The lines should not cross and should begin at the surfaces of the spheres. The shape of the field lines should demonstrate how they spread outwards and the repulsive nature of the fields from the two spheres.

To calculate the electric field at point P due to sphere A and sphere B, we use the formula for the electric field due to a point charge:

$E = \frac{kQ}{r^2}$

Where:

• $k = 9 \times 10^9 \ \text{N m}^2/\text{C}^2$,
• $Q = 5 \times 10^{-6} \ \text{C}$,
• $r$ is the distance from the charge to point P.

For sphere A, the distance is $1.25 \text{ m}$:

$E_{PA} = \frac{(9 \times 10^9)(5 \times 10^{-6})}{(1.25)^2} = 2.88 \times 10^4 \ \text{N/C} \text{ to the right}$

For sphere B, the distance is $0.75 \text{ m}$:

$E_{PB} = \frac{(9 \times 10^9)(5 \times 10^{-6})}{(0.75)^2} = 8.00 \times 10^4 \ \text{N/C} \text{ to the left}$

The net electric field at point P is the vector sum of $E_{PA}$ and $E_{PB}$. Since they are in opposite directions:

$E_{net} = E_{PB} - E_{PA} = 8.00 \times 10^4 \ \text{N/C} - 2.88 \times 10^4 \ \text{N/C} = 5.12 \times 10^4 \ \text{N/C} \text{ to the left}$