Solution
Let the original charge be $Q$. We are given data for point A and B.
At Point A:
$$V_A = \frac{kQ}{r_A} = 7 \, \text{V}$$ $$E_A = \frac{kQ}{r_A^2} = 3 \, \text{V/m}$$Dividing these gives the distance $r_A$:
$$r_A = \frac{V_A}{E_A} = \frac{7}{3} \, \text{m}$$Substituting back to find $kQ$:
$$kQ = E_A r_A^2 = 3 \times \left(\frac{7}{3}\right)^2 = \frac{49}{3}$$At Point B (New Condition):
The charge magnitude is tripled, so $Q’ = 3Q$. We are told the new electric field at B is $3 \, \text{V/m}$.
Substituting $kQ = 49/3$:
$$\frac{49/3}{r_B^2} = 1 \implies r_B = \sqrt{\frac{49}{3}} = \frac{7}{\sqrt{3}}$$Calculate New Potential at B ($V’_B$):
$$V’_B = \frac{k(3Q)}{r_B} = 3 \frac{kQ}{r_B}$$ $$V’_B = 3 \times \frac{49/3}{7/\sqrt{3}} = 49 \times \frac{\sqrt{3}}{7} = 7\sqrt{3}$$Using $\sqrt{3} \approx 1.732$:
$$V’_B \approx 7 \times 1.732 = 12.124 \, \text{V}$$The value is closest to 12 V.
Answer: (b) 12 V