Question

What is the ratio ef/ei of the final electric field strength ef to the initial electric field strength ei if q is doubled?

Answer 1

When the charge q is doubled, the final electric field strength ef is twice as strong as the initial electric field strength ei, resulting in a ratio ef/ei of 2:1.

Step-by-step explanation:

The electric field (E) created by a point charge (q) at a distance (r) is given by Coulomb's law, which states that E = k * q / r^2, where k is Coulomb's constant.

If the charge q is doubled to become 2q, the new electric field strength ef at the same distance will be Ef = k * (2q) / r^2.

This means that the final electric field strength will be twice as strong as the initial electric field strength because Ef = 2 * (k * q / r^2) = 2 * Ei.

Therefore, the ratio ef/ei of the final electric field strength to the initial electric field strength when q is doubled is 2:1.

Answer 2

As we know that electric field at a point due to a point charge is given by

`E = (kq)/(r^2)`

here we can see that electric field is directly depends on the magnitude of charge

so if we change the magnitude of charge by some factor then electric field will change by same factor

so here as given that charge is doubled

then electric field due to that charge at the same point also get doubled