A wire is bent in th eform of a regular hexagon and a total cahrge q is distributed uniformly on it. What is the electric field at the centre? you may answer this part without making any numerical calculations.

Open in App

Solution

Since it is a regular hexagon. So, it forms an equipotential surface. Hence the charge at each point is equal hence the net electric field at the centre is 0.

Suggest Corrections

Similar questions

Q. A wire is bent in the form of a regular hexagon and a total charge q is distributed uniformly on it. What is the electric field at the centre? You may answer this part without making any numerical calculations.

Q. A wire is bent in the form of a regular hexagon and a total charge

q

is distributed uniformly on it. What is the electric field at the centre ?

Q. Charge Q is distributed uniformly on length

l

of a wire. It is bent in the form of a semicircular ring. Find the electric field at the centre of the ring.

A rod of length L has a total charge Q distributed uniformly along its length. It is bent in th shape of a semicircle. Find the magnitude of the electric field at the centre of curvatere of he semicircle.

Q. A rod of length

L

as a total charge

Q

distributed uniformly along its length. It is bent in the shape of a semicircle. Find the magnitude of the electric field at the centre of curvature of the semicircle.

Join BYJU'S Learning Program