On a planet, the acceleration due to gravity is the same as that on Earth but th

Question

On a planet, the acceleration due to gravity is the same as that on Earth but there is a strong downward electric field with the field being uniform close to the planet

Explanation / Answer

v = 0 = at top = 20.1 - [g + qE/m]*t1 t1 = time to top = 20.1/[g +qE/m] -------- (A) height gained = h = [20.1^2 -0]/2[g + qE/m] --- (A1) ------------ now it falls from REST for (h) height taking (t2) time acceleration = +[g + qE/m] h = o + 0.5 [g + qE/m] t2^2 >>>>>> using (A1) [20.1^2]/2[g + qE/m] = 0.5 [g + qE/m] t2^2 t2^2 = [20.1^2]/[g + qE/m]^2 t2 = 20.1/[g + qE/m] ----------- (A2) ================ t1+t2 = 4.10 >>>>>> given 20.1/[g + qE/m] = 2.05 [g + qE/m] = 20.1/2.05 >>>>>>>> (A3) qE/m = 9.804878 - 9.8 = 4.878*10^-3 E = 4.878*10^-3*2 /5*10^-6 = 1951.2 N/C >>>>>> ============================== h = [20.1^2 -0]/2[g + qE/m] putting value from (A3) h = [20.1^2]/2*{20.1/2.05} h = 20.1*2.05/2 = 20.6 m ================= potential difference dV = E * separation = E * h dV = 1951.2 * 20.6 = 40194.72 Volt

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