On an imaginary planet the acceleration due to gravity is same as that on Earth but there is also a downward electric field that is uniform close to the planet's surface. A ball of mass m carrying a charge q is thrown upward at a speed v and hits the ground after an interval t, What is the magnitude of potential difference between the starting point and top point of the trajectory?

Options

A. `(mv)/(2q)(v-(g t)/2)`
B. `(mv)/q(v-(g t)/2)`
C. `(mv)/(2q)(v-g t)`
D. `(2mv)/q(v-g t)`

Answer

A

Explanation

Net downwards aceleration on body of mass `m=(g+(qE)/m) = a_(net)`
ifE = uniform electric field in downwards direction
If it hits after time `t=(2v)/[g+(qE)/m] E downarrow downarrow downarrow downarrow /planet`
at maximum height `v_(f)=0`
`v_(f)^(2)=v_(i)^(2)-2a_(net)h` Rightarrow in uniform field (Gravitation + Electric field time to reach highest point =`t//2`]
`v^(2)=2[g+(qE)/m]h`
`DeltaV` between ground and highest point
`DeltaV =(E)h`
`a_(net) = (2v)/tRightarrow (g+(qE)/m)=(2v)/t`
`q/m E =((2v)/t-g)`
`E=m/q((2v)/t-g)and h=(average velocity)xxt`
`h=v/2xxt`
`so DeltaV = m/q ((2v)/t-g)(v/2t)`
`=(mv)/(2q)(v-(g t)/2)`