Two point charges, A and B, each with a charge of +3 x 10⁻⁹ C, are stationary on a horizontal surface - NSC Physical Sciences - Question 7 - 2023 - Paper 1

The electric field lines due to both charges A and B will radiate outward since they are both positive charges. The lines will begin at each charge and extend towards point P. The resultant electric field at point P will be the vector sum of the electric fields due to each charge.

To find the value of r, we can use the formula for the electric field due to a point charge:

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

The electric field at point P due to charge A (at distance r) is:

$E_A = k \frac{3 \times 10^{-9}}{r^2}$

The electric field at point P due to charge B (at distance 2r) is:

$E_B = k \frac{3 \times 10^{-9}}{(2r)^2} = k \frac{3 \times 10^{-9}}{4r^2}$

The net electric field at point P is:

$E_{net} = E_A - E_B$

Given that the magnitude of the net electric field is 27 N·C⁻¹, we substitute:

$27 = k \frac{3 \times 10^{-9}}{r^2} - k \frac{3 \times 10^{-9}}{4r^2}$

Combining these gives:

$27 = k \left( \frac{3 \times 10^{-9}}{r^2} - \frac{3 \times 10^{-9}}{4r^2} \right)$

This simplifies to:

$27 = k \frac{3 \times 10^{-9} \cdot 4 - 3 \times 10^{-9}}{4r^2} = k \frac{9 \times 10^{-9}}{4r^2}$

Thus, we have:

$27 imes 4r^2 = k \times 9 \times 10^{-9}$

Solving for r provides:

$r = 0.87 \, m$

The force experienced by an electron in an electric field is given by:

$F = E \cdot q$

Where:

• F is the force,
• E is the electric field strength (27 N·C⁻¹),
• q is the charge of the electron (approximately $-1.6 \times 10^{-19}$ C).

Substituting these values:

$F = 27 imes (-1.6 \times 10^{-19})$

Calculating this gives:

$F = -4.32 \times 10^{-18} \, N$

Since we need the magnitude, the answer is:

$4.32 \times 10^{-18} \, N$.