What is the magnitude of the electric field at point P, located at (4.50 cm, 0),

Question

What is the magnitude of the electric field at point P, located at (4.50 cm, 0), due to Q1 alone?
9.22×106 N/C

What is the x-component of the total electric field at P?
2.46×107 N/C

What is the y-component of the total electric field at P?
-6.14×106 N/C

What is the magnitude of the total electric field at P?
2.53×107 N/C

Now let Q2 = Q1 = 3.00 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?

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

Two charges, Q1= 3.00 C, and Q2= 6.60 C are located at points (0,-3.00 cm ) and (0,+3.00 cm), as shown in the figure.

Explanation / Answer

if both the charges are positive and located at (0,3 ),(0,-3) then electric field due to both charges will add up in horizontal direction while vertical component will cancel each other.

suppose electric field makes angle a with the horizontal then the net electric field at point P can be calculated as -

E = 2 E' cos a

E = 2*9.23 cos 33.69

E = 15.35 N/C

MAGNITUDE OF FORCE ON AN ELECTRON -

F = eE

F = 2.45*10-18 N

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