What is the magnitude of the electric field at point P, located at (6.50 cm, 0),

Question

What is the magnitude of the electric field at point P, located at (6.50 cm, 0), due to Q1 alone?

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 = 3.90 ?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?

Explanation / Answer

i answered just a similar question, please take help from it if possible,

Q1 creates a field that is up and to the right at angle of arctan(3.5/4.5) =37.87o

Q2 creates a field that is down and to the right at angle of arctan(-3.5/4.5) =37.87o

Now E = k*q/r^2 The distance squared for both is 0.035^2 + 0.045^2 = 0.00325

Ex = k*Q1/r^2*cos(37.87) + k*Q2/r^2*cos(-37.87)

= 9.0x10^9*2.00x10^-6/0.00325*cos(37.87) + 9.0x10^9*6.30x10^-6/0.00325*cos(-37.87)

=1.81x10^7N/C

and

b) Ey = k*Q1/r^2*sin(37.87) + k*Q2/r^2*sin(-37.87)

= 9.0x10^9*2.00x10^-6/0.00325*sin(37.87) + 9.0x10^9*6.30x10^-6/0.00325*sin(-37.87)

= -7.31x10^6N/C


c) mag = sqrt(Ex^2 + Ey^2) = sqrt((1.81x10^7)^2 + (7.31x10^6)^2) =1.96x10^7N/C

d) Now only the y component is non zero due to symmetry

So Ey = 2*k*Q/r^2*sin() = 2*9.0x10^9*2.0x10^-6/0.00325*sin(37.87) = 6.80x10^6N/C

e) F = E*q = 6.80x10^6N/C*1.60x10^-19C = 1.09x10^-12N

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.