a) Four identical particles with charge 3 μμC are located at the corners of a square. The square's sides are 2 cm in length. A fifth particle of charge 6 μμC is placed directly at the center of the square. What is the magnitude of the total electric force on any one of the four particles at the corners?
b) If you could adjust the charge of the particle at the center, what should you choose so that the net electric force on any of the corner particles is zero?
From Coulomb's Law
Where k is coulomb constant = 9x109 and r is the distance between charges q1 and q2
Lets find Force on charge at corner B
Force due to charge 3uC at corner A , FA = 9*109 * 3*10-6 *3*10-6 / 0.022
FA = 202.5 N (+ve x direction)
Force due to charge 3uC at corner C , FC = 9*109 * 3*10-6 *3*10-6 / 0.022
FC = 202.5 N (+ve y direction)
Distance from corner D to B = *0.02 = 0.0283 m
Force due to charge 3uC at corner D , FD = 9*109 * 3*10-6 *3*10-6 / 0.02832
FD = 101.137 N (at angle 45 deg to x axis)
center E is in the middle of diagonal DB ,so r = 0.0141 m
Force due to charge 6uC at centerE , FE = 9*109 * 6*10-6 *3*10-6 / 0.01412
FE = 814.85 N (at angle 45 deg to x axis)
Split FE and FD to its components and ADD
Force on x direction
Fx = FA + FDcos45 + FEcos45 ---------------------(1)
= 202.5 + 101.137* cos45 + 814.85cos45
= 850.2 N
Fy = FC + FDsin45 + FEsin45
=202.5 + 101.137* sin45 + 814.85sin45
= 850.2 N
total force F = = 1202.36 N
ANSWER : F = 1202.36 N
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(b)electric force on corner particles to zero
from equation (1)
if Fx = FA + FDcos45 - FEcos45 =0 horizontal force is balanced
find FE everythg else is unchanged
FE = (FA + FDcos45 )/cos45
= (202.5 + 101.137* cos45)/cos45
FE=387.51 N
k*3*10-6 * X /r2 = 387.51 N here r =a/ =0.0141 m
X*9*109 *3*10-6 /(0.0141)2 =387.51 N
X = 387.51 *(0.0141)2 /27000
X= -2.853uC
ANSWER : change 6uC to -2.853uC to reduce net electric force on all corner particles to ZERO
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