A ball of mass 100g and having a charge of `4.9xx 10^(-5)` C is released from rest in a region where a horizontal electric field of `2.0 xx 10^4 NC^(-1)` exists. (a) find the resultant force acting on the ball. (b) What will be the path of the ball ? (c ) Where will the ball be at the end of 2 s?

Explanation

`M=100gm, `
` q=4.9xx10^-5, `
` F_g=mg, `
` F_e=eq, `
` e=2xx10^4N//C `
` So, the particle moves due to the resultant R `
` R^2=F_g^2+F_e^2 `
` = (0.1xx9.8)^2+(4.9xx10^-5xx2xx10^4)^2 `
` =0.9604+96.04xx10^-2 `
` 1.9208 So, R=1.3859N `
` tan theta=1 `
` So, theta=45^@ `
Hence path is straight along resultant force at an angle `45^@`with horizontal Displacement (verticle)
` 1/2xx9.8xx2xx2=19.6m `
Displacement (horizontal)
` =s=1/2xxaxxt^2 `
`1/2xxt^2xx2=19.6m `
Net displacement
` =sqrt((19.6)^2+(19.6)^2) `
` =sqrt(768.932)=27.7m`