7.1 Define the term electric field at a point - NSC Physical Sciences - Question 7 - 2019 - Paper 1

Charge q2 is positive. The electric field due to q1, which is negative, points to the right. Thus, for the net field to be zero at point P, the electric field due to q2 must be directed to the left, indicating that q2 is positive.

The electric field due to q1 is given by: $E_1 = \frac{k |q_1|}{r^2}$ At point P, the electric field due to q2 must equal that of q1 when in equilibrium: $E_2 = \frac{k |q_2|}{r^2}$ Since these fields are equal in magnitude but opposite in direction: $E_1 = E_2$ Substituting gives: $\frac{9 \times 10^9 \times 3 \times 10^{-9}}{(0.1)^2} = \frac{9 \times 10^9 \times |q_2|}{(0.3)^2}$ Solving gives: $|q_2| = 4.48 \times 10^{-8} \, C$

Coulomb’s law states that the electrostatic force of attraction or repulsion between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

Using Coulomb’s law, the force F between charges q1 and q2 is given by: $F = \frac{k |q_1 q_2|}{r^2}$ Substituting the known values: $F = \frac{(9 \times 10^9)(3 \times 10^{-9})(4.48 \times 10^{-8})}{(0.4)^2}$ This calculation gives: $F = 1.44 \times 10^{-5} \, N$

Yes, the diagram is correct. After the two charges are brought into contact, they equalize their charge, becoming both equal and positive, which is illustrated correctly in the diagram.