Question

You are given two unknown point charges, Q1, and Q2. At a point on the line joining them, one third of the way from Q1 to Q2 the electric field is zero. What can you say about these two charges?

Answer 1

Answer:

• The charges are of the same sign

• The charge Q2 its four times the charge Q1

Explanation:

• The charges are of the same sign.

If they were of opposite signs, in the line joining them, the electric field will not be zero at any point, as in all points of this line will be pointing towards the negative charge.

• The charge Q2 its four times the charge Q1.

The equation for the electric field produced by a point charge its:

$E = \frac{1}{4 \pi \epsilon_0} \frac{Q}{d^2}$

taking in consideration that the electric field its a vector that points radially outward from the point charge, and the equation above only gives us its length.

Now, If we got a zero electric field, that must mean that the electric fields from both charges are, in magnitude, equals. This means:

$\frac{1}{4 \pi \epsilon_0} \frac{Q_1}{d_1^2}=\frac{1}{4 \pi \epsilon_0} \frac{Q_2}{d_2^2}$

$\frac{Q_1}{d_1^2} = \frac{Q_2}{d_2^2}$

We also now that the distance to Q1 from this point its one third the distance from Q1 to Q2, this must mean that $d_2$ its 2 times $d_1$.

$\frac{Q_1}{d_1^2} = \frac{Q_2}{ (2 * d_1)^2}$

We can get ride of the $d_1$ factor that appears on both sides:

$\frac{Q_1}{d_1^2} = \frac{Q_2}{ 2^2 d_1^2}$

$\frac{1}{d_1^2} Q_1 = \frac{1}{d_1^2}\frac{Q_2}{ 4 }$

$Q_1 = \frac{Q_2}{ 4 }$

And we finally get:

$Q_2 = 4 * Q_1$

Answer 2

The formula that is applicable here is E = kQ/r^2 in which the energy of attraction is proportional to the charges and inversely proportional to the square of the distance. In this case, 
kQ1/(r1)^2 = kQ2/(r2)^2  r1=l/3, r2=2l/3solve Q1/Q2
kQ1/(l/3)^2 = kQ2/(2l/3)^2  kQ1/(l^2/9) = kQ2/(4l^2/9)Q1/Q2 =   1/4