Two point charges +4q and +q are placed 30 cm apart. At what point on the line joining them is the electric field zero?
(a) 15 cm from charge 4q
(b) 20 cm from charge 4q
(c) 7.5 cm from charge q
(d) 5 cm from charge q
-4.7cm
To determine the point on the line joining the charges where the electric field is zero, we need to consider the electric field contributions from each charge separately.
First, let's consider the electric field due to the +4q charge. The electric field at a point on the line is given by:
E1 = k * (4q) / r1^2
Where k is the Coulomb's constant, q is the charge, and r1 is the distance from the point to the +4q charge.
Similarly, for the +q charge, the electric field at a point on the line is given by:
E2 = k * q / r2^2
Where r2 is the distance from the point to the +q charge.
Since the electric field is a vector quantity, it will be zero if the magnitudes of E1 and E2 are equal, but in opposite directions.
Let's assign the point we are interested in to be at a distance x from the +4q charge, and (30 - x) from the +q charge.
Therefore, we have:
E1 = E2
k * (4q) / x^2 = k * q / (30 - x)^2
Simplifying this equation, we can cancel out the k and q terms and rearrange:
4 / x^2 = 1 / (30 - x)^2
Cross multiplying and simplifying further:
4(30 - x)^2 = x^2
1200 - 80x + x^2 = x^2
1200 - 80x = 0
80x = 1200
x = 1200 / 80
x = 15 cm
Therefore, the point where the electric field is zero is 15 cm from the +4q charge.
The correct answer is (a) 15 cm from charge 4q.