12. The distance between the two charges, which are $25 \mathrm{mC}$ and $36 \mathrm{mC}$, is * 2 points $11 \mathrm{~cm}$. At what point on the line joining the two will the electric field strength will be zero?
A distance of $5 \mathrm{~cm}$ from $25 \mathrm{mC}$ charge
A distance of $5 \mathrm{~cm}$ from $36 \mathrm{mc}$ charge
A distance of $3 \mathrm{~cm}$ from $25 \mathrm{mC}$ charge
A distance of $3 \mathrm{~cm}$ from $36 \mathrm{mc}$ charge

Step 1 ：Given two charges, Q1 = 25 mC and Q2 = 36 mC, and the distance between them is 11 cm. We need to find the point on the line joining the two where the electric field strength will be zero.

Step 2 ：The electric field strength will be zero at a point where the electric fields due to the two charges cancel each other out. This can happen at two possible locations - one between the charges and one outside the charges.

Step 3 ：The electric field E due to a point charge Q at a distance r is given by Coulomb's law: \(E = kQ/r^2\), where k is Coulomb's constant (9 x 10^9 N.m^2/C^2).

Step 4 ：We can set up two equations based on the two possible locations of the point where the electric field strength is zero. Let's denote the distances from the point to the charges as r1 and r2.

Step 5 ：For the point between the charges, we have: \(E1 = E2\), which gives us \(kQ1/r1^2 = kQ2/(11-r1)^2\).

Step 6 ：For the point outside the charges, we have: \(E1 = -E2\), which gives us \(kQ1/r2^2 = -kQ2/(11+r2)^2\).

Step 7 ：Solving these equations, we find that for r1, we have two solutions: -0.55 m and 0.05 m. The negative solution doesn't make sense in this context, so we discard it. The positive solution, 0.05 m, corresponds to a distance of 5 cm, which is one of the options in the question. This is the point between the charges where the electric field strength is zero.

Step 8 ：For r2, we have complex solutions, which also don't make sense in this context. So, there is no point outside the charges where the electric field strength is zero.

Step 9 ：Final Answer: The electric field strength will be zero at a point \(\boxed{5 \mathrm{~cm}}\) from the $25 \mathrm{mC}$ charge.