Two charges, Q 1 = 2.30 C, and Q 2 = 5.50 C are located at points (0,-3.00 cm )

ID: 1630423 • Letter: T

Question

Two charges, Q1= 2.30 C, and Q2= 5.50 C are located at points (0,-3.00 cm ) and (0,+3.00 cm), as shown in the figure.

What is the magnitude of the electric field at point P, located at (6.00 cm, 0), due to Q1 alone?
4.59×106 N/C

What is the x-component of the total electric field at P?
1.39×107 N/C

What is the y-component 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?

Now let Q2 =Q1 = 2.30 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?

Two charges, Q1= 2.30 C, and Q2= 5.50 C are located at points (0,-3.00 cm ) and (0,+3.00 cm), as shown in the figure.

1)The electric field because of q1 alone is:

E1 = k q1/r1^2

E1 = 9 x 10^9 x 2.3 x 10^-6 / (0.06^2 + 0.03^2) = 4.6 x 10^6 N/C

Hence, E1 = 4.6 x 10^6 N

2)theta = tan-1 (3/6) = 26.57 deg

Field due to Q2 is;

E2 = k q2/r2^2

E2 = 9 x 10^9 x 5.5 x 10^-6 / (0.06^2 + 0.03^2) = 1.1 x 10^7 N/C

So the X component will be:

Ex = E1 cos(theta) + E2 cos(theta)

Ex = 4.6 x 10^6 x cos26.57 + 1.1 x 10^7 x cos26.57 = 1.9 x 10^7 N/C

Hence, Ex = 1.9 x 10^7 N/C

3)Ey = E1 sin(theta) - E2 sin(theta)

Ey = 4.6 x 10^6 x sin(26.57) - 1.1 x 10^7 x sin(26.57) = -6.34 x 10^6 N/C

Hence, Ey = -6.34 x 10^6 N/C

4)we know that, force on a charged particle in an electric field is:

F = q E

E = sqrt [Ex^2 + Ey^2]

E = sqrt [(1.9 x 10^7)^2 + (-6.34 x 10^6)^2] = 2 x 10^7 N/C

F = 1.6 x 10^-19 x 2 x 10^7 = 3.2 x 10^-12 N

Hence, F = 3.2 x 10^-12 N

5)If Q1 = Q2 = 2.3 uC, E1 = E2 = E, so the y component of the field cancels out and x component sums to:

Ex = E(tot) = 2 E cos(theta)

Ex = 2 x 4.6 x 10^6 x cos26.57 = 1.23 x 10^6 N/C

Hence, E(tot) = 1.23 x 10^6 N/C

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