Two charges Q1 240 C and Q2 500 C are located at points 0400

Two charges, Q1= 2.40 C, and Q2= 5.00 C are located at points (0,-4.00 cm ) and (0,+4.00 cm), as shown in the figure. Answer questions below in clear and exact answers (and convert necessary parts to meters)

1. What is the magnitude of the electric field at point P, located at (4.50 cm, 0), due to Q₁ alone?

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

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

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

5. Now let Q₂ = Q₁ = 2.40 C. 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?

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

Solution

Here, two charges are there at (-4, 0 ) and (4,0), namely Q1 and Q2 respectively. [Q1 is the one at (-4,0)]

1.) Now magnitude of electric field at (4.5,0) due to Q will be = E = kQ1 / r^2

E = 8.99 x 10^9 x 2.4 x 10^-2 / 36.25 = 5.952 x 10^6 N/C

2.) As above the electric field at P due to Q2 = 8.99 x 10^9 x 5 x 10^-2 / 36.25 =  12.4 x 10^6 N/C

Now we need to take the x component of the electric fields due to Q1 and Q2

Product of electric field and cosine of angle the vectors make with x axis will give the x component of the electric field.

Cos(Angle with x axis) = 4.5 / 6.02 = 0.7475

Therfore, net electic field towards the x direction = Ex = (5.952 + 12.4) x 0.7475 x 10^6 = 13.718 x 10^6 N/C

3.) Now, the sine(Angle with x axis) = 4/6.02 = 0.66445

However, for the y axis, the direction of the electric fields are opposite for both the charges, hence the net electric field along the y direction will be the difference of the electric fields due to individual charges.

Ey = (5.952 - 12.4) x 0.66445 x 10^6 = -4.2844 x 10^6 N/C

4.) Magnitude of total electric field = sqrt (Ex^2 + Ey^2) = 10^6 x sqrt(13.718^2 + 4.2844^2) = 14.37 x 10^6 N/C

5.) As the charges are same now, both are equal to Q1, the net electric field along the y axis will cancel out. Further, the net electric field towards x axis will be twice of what we have due to Q1 in part 2

Therefore net, electric field will be same as that along x axis = 2 x 13.718 x 10^6 = 27.436 x 10^6 N/C

6.) Magnitude of force on an electron = qE = 1.6 x 10^-19 x 27.436 x 10^6 = 43.8976 x 10^-13 N