Question

Two charges, Q1=2.90 μC, and Q2=5.30 μC are located at points (0,-2.00 cm ) and (0,+2.00 cm), as shown in the figure. 

What is the magnitude of the electric field at point P, located at (5.00 cm, 0), due to Q1 alone? 

The electric field at position P due to charge Q1 is not influenced by charge Q2. Therefore, ignore charge Q2 and apply Coulomb's Law. Remember to convert all units to the SI unit system. 

What is the x-component of the total electric field at P? 

What is the y-component of the total electric field at P? 

What is the magnitude of the total electric field at P? 

Now let Q2 = Q1 = 2.90 μC. Note that the problem now has a symmetry that you should exploit in your solution. What is the magnitude of the total electric field at P? 

Given the symmetric situation of the previous problem, what is the magnitude of the force on an electron placed at point P? 

Answer 1

Answer :

1) Required E(magnitude) = [(9*10^9)*(2.90*10^-6)]/ [(2^2 +5.0^2)*10^-4]
E = (26.1*10^3)/(29*10^-4) = 9*10^6 N/C
E will make an angle of tan^-1[2/5.0] = 21.80 degree with positive x-axis

2)X component, TEx of Total electric field = [{(9*10^9)*(5.30*10^-6)}/[{(4^2 +6.50^2)*10^-4}] cos 21.80 +
(9*10^6)*cos 21.80 or
TEx = [(54*10^3)/(29*10^-4) + (9*10^6)]*cos 21.80 = [(16.48+9)*cos 21.80 ]*10^6 N = 23.62*10^6 N/C
TEx will be along positive x axis

3)Similarly TEy = {16.48 - 9]*sin 21.80 = 2.76*10^6 N/C
Direction of TEy will be along negative y direction

4)Total electric field, TE = (10^6)*sq rt[23.62^2 + 2.76^2] = 23.78*10^6 N/C
Direction of TE will make an angle of tan^-1[2.76/23.62] = 6.62 degree with positive x-axis in clockwise direction. vector lies in 4th quadrant, where x-component is positive and y-component is negative

5)TE(symmetric) = 2*(9*10^6)*cos 31.6 = 16.71*10^6 N/C
[y components cancel because of symmetry and contribution of x components are equal from both charges]

6)Required force = (16.71*10^6)*(1.6*10^-19) = 2.67*10^-12 N along x-axis towards origin.