A planet travels in an elliptical orbit about a star as shown. At what pair of points is the speed of the planet the same?

Options

A. `kq//a^(2)` at an angle `45^(@)` above the +x-axis
B. `kq//a^(2)` at an angle `45^(@)` below the -x-axis
C. `3kq//a^(2)` at an angle `45^(@)` above the -x-axis
D. `3kq//a^(2)` at an angle `45^(@)` below the +x-axis

Answer

B

Explanation

At center electric due to both 3q will be canceled. Now net field at center is
`E=E_(1)-E_(2)`
`=(k2q)/((a^(2)))-(kq)/((a)^(2))=kq/((a)^(2))`
This is `45^(@)` below negative x-axis
.