Two point charges +4q and +q are kept at a distance of 30 cm from each other. At which point between them, the field intensity will be equal to zero?
A. 15cm away from the +4q charge
B. 20cm away from the +4q charge
C. 7.5cm away from the +q charge
D. 5cm away from the +q charge
Answer:B
Two point charges +4q and +q are kept at a distance of 30 cm from each other. At which point between them, the field intensity will be equal to zero?
A. 15cm away from the +4q charge B. 20cm away from the +4q charge C. 7.5cm away from the +q charge D. 5cm away from the +q charge
Explanation: The electric field at a distance of r from a charge q is equal to \(\frac {q}{4\pi\varepsilon_or^2}\). Let the electric field intensity will be zero at a distance of x cm from +4q charge, so the fields due to the two charges will balance each other at that point. Therefore \(\frac {4q}{4\pi\varepsilon_ox^2}=\frac {q}{4\pi\varepsilon_o(30-x)^2}\). Solving this we get x=20cm. Therefore the point will be 20cm away from the +4q charge.