HELP WITH C PLEASE Problem 20.60 MC Two 1.5-cm-diameter disks face each other, 1

Question

HELP WITH C PLEASE

Problem 20.60 MC Two 1.5-cm-diameter disks face each other, 1.0 mm apart. They are charged to t 15 nC Part B What is the magntiude of the electric force acting on a proton due to the electric field between the discs? Express your answer using four significant figures. 535x10 2 N Submit Hints My Answers Give Up Review Part Correct Part C A proton is shot from the negative disk toward the positive disk. What launch speed must the proton have to just barely reach the positive disk? Assume the weight of the proton is negligible. Express your answer using three significant figures. m/s

Explanation / Answer

The electric field between the plates for the first part as given
So E = 9.591x10^6N/C

now, Using conservation of energy :

(K + U) b = (K + U)t...Here Ut - Ub = q*(Vt - Vb)...and Vt - Vb = E*d = 9.591x10^6V/m x 1.0x10^-3m = 9591V

Since Kt = 0 we have Kb = q*E*d = 1.60x10^-19C x 9591 = 1.53456x10^-15J

So 1/2*m*v^2 = 1.53456x10^-15J
or, v = sqrt(2x1.53456x10^-15J/m) = sqrt(2x1.53456x10^-15J/1.67x10^-27kg) = 1.3556x10^6m/s

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