99 Percentile Qs Bank for JEE MainPhysicsElectrostatics
A cube of side L has point charges +q located at its seven vertices and -q at remaining one vertex. The electric field at its centre is found to be | E |= ( q 4 ₀ L^2 ) . The magnitude of constant is
Options
- A4 3
- B8 3
- C3
- D1
Correct answer
B. 8 3
Step-by-step solution
The given situation is shown in the following figure Clearly, 5 x -components of E will be left to right, while 3 x -components of E will be from right to left. From the figure given below r^2= L^2 4 + L^2 2 = 3 L^2 4 So, E due to one point charge |E|= K q r^2 = 4 3 K q L^2 Electric field at the centre of cube due to contribution of all charges is given as |E|_ Total = 4 3 K q r^2 (5-3)= 8 3 K q r^2 = 8 3 ( q 4 ₀ L^2 ) Given, |E|= [ q 4 ₀ L^2 ] [ r^2= 3 L^2 4 ] Clearly, = 8 3