Question

Please explain how you approach this so I can emulate your problem solving strategies on like problems

Suppose we had a large, positively charged plate in an upright (vertical) position and a point charge with positive charge +Q. In lecture, you saw that a large charged plate gives rise to an electric field which is constant in space: we will call this electric field Eplate. Let us suppose also that we had a s bl with positive charge +q and mass m. The situation is depicted in Figure 1. +q,m +Q Figure 1 The objective of this problem is to find the location of the ball with charge +q and mass n such that it remains in place, even with electrical and gravitational forces acting Hint: This location is unique Part a Draw a free body diagran for the charged bal and clearly label the forces acting on the ball.

Answer 1

Part A)

Part A) Part B) r = distance of ball from the charge + Q theta = angle of the line joining the charge + Q with the ball from the horizontal or angle of force of repulsion from horizontal x-direction using Pythagorean theorem r = squareroot (x^2 + y^2) also, tan theta = y/x so theta = tan^-1(y/x) Cos theta = x/r = x/sqrt(x^2 + y^2) and Sin theta = y/r = y/sqrt(x^2 + y^2) force of repulsion by the charge + Q on the ball is given as F_r = k Q q/r^2 F_r = k Q q/(sqrt(x^2 + y^2))^2 F_r = k Q q/(x^2 + y^2) eq^-1 Net force along the vertical direction or Y-direction is given as F_y = F_r Sin theta - mg using eq^-1 F_y = k Q q Sin theta/(x^2 + y^2) F_plate F_x = (k Q q/(x^2 + y^2)) (y/sqrt(x^2 + y^2)) F_plate F_x = k Q q y/(x^2 + y^2)3/2 - F_plate

part B)

r = distance of ball from the charge +Q

$\theta$ = angle of the line joining the charge +Q with the ball from the horizontal or angle of force of repulsion from horizontal x-direction

using Pythagorean theorem

r = sqrt(x² + y²)

also , tan$\theta$ = y/x

so $\theta$ = tan^-1(y/x)

Cos$\theta$ = x/r = x/sqrt(x² + y²)     and Sin$\theta$ = y/r = y/sqrt(x² + y²)

force of repulsion by the charge +Q on the ball is given as

F_r = k Q q/r²

F_r = k Q q/( sqrt(x² + y²))²

F_r = k Q q/(x² + y²)                                eq-1

Net force along the vertical direction or Y-direction is given as

F_y = F_r Sin$\theta$ - mg

using eq-1

F_y = k Q qSin$\theta$/(x² + y²) - mg                                 

F_y = (k Q q/(x² + y²)) (y/sqrt(x² + y²)) - mg

F_y = k Q q y/(x² + y²)^3/2 - mg                                  eq-2

Net force along the horizontal direction or X-direction is given as

F_x = F_r Cos$\theta$ - F_plate

using eq-1

F_x = k Q q Cos$\theta$/(x² + y²) - F_plate                               

F_x = (k Q q/(x² + y²)) (x/sqrt(x² + y²)) - F_plate    

F_x = k Q q x/(x² + y²)^3/2 - F_plate                  eq-3         

c)

at equilibrium

F_y = 0

using eq-2

k Q q y/(x² + y²)^3/2 - mg = 0

k Q q y/(x² + y²)^3/2 = mg

y/(x² + y²)^3/2 = mg/(k Q q)                               eq-4

also , F_x = 0

using eq-3

k Q q x/(x² + y²)^3/2 - F_plate = 0

x/(x² + y²)^3/2 = F_plate /(k Q q)                       eq-5

dividing eq-5 by eq-4

x/y = F_plate /(mg)

x = F_plate y /(mg)                                              eq-6

using eq-5 and eq-6

(F_plate y /(mg))/((F_plate y /(mg))² + y²)^3/2 = F_plate /(k Q q)

Since F_plate = q E_plate

hence

y = sqrt((kQq/(mg)) (1 + (q E_plate/(mg))²)^-3/2)

using eq-4

y/(x² + y²)^3/2 = mg/(k Q q)

(sqrt((kQq/(mg)) (1 + (q E_plate/(mg))²)^-3/2))/(x² + ((kQq/(mg)) (1 + (q E_plate/(mg))²)^-3/2))^3/2 = mg/(k Q q)

x = sqrt((kQ/E_plate) (1 + (mg/(q E_plate))²)^-3/2)

D)

x = sqrt((kQ/E_plate) (1 + (mg/(q E_plate))²)^-3/2)

inserting the values

x = sqrt(((8.99 x 10⁹)(8.50 x 10^-3)/(7 x 10⁴)) (1 + ((0.010 x 9.81)/((3 x 10^-6) (7 x 10⁴)))²)^-3/2)

x = 28 m

y = sqrt((kQq/(mg)) (1 + (q E_plate/(mg))²)^-3/2)

inserting the values

y = sqrt(((8.99 x 10⁹)(8.50 x 10^-3) (3 x 10^-6)/(0.010 x 9.8)) (1 + ((3 x 10^-6) (7 x 10⁴)/(0.010 x 9.81))²)^-3/2)

y = 13 m