Two charges, Q1= 2.90 ?C, and Q2= 5.30 ?C are located at points (0,-4.00 cm ) an

Question

Two charges, Q1= 2.90 ?C, and Q2= 5.30 ?C are located at points (0,-4.00 cm ) and (0,+4.00 cm), as shown in the figure.

What is the magnitude of the electric field at point P, located at (5.50 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 ?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

due to Q1 alone, E1 = k*Q1/D^2 ; D = sqrt(0.04^2 + 0.055^2) = = 9*10^9*2.9*10^-6 / (0.068*0.068) = 5.644*10^6 N/C angle made by electric field by +ve x axis = 36 up (due to Q1) E2 due to Q2 = 10.315 * 10^6 N/C (angle = same 36 down with + x axis) Total electric field (x component) = (E1+E2) cos 36 = 12.911*10^6 N/C in +ve x direction. Y component = (E1-E2)*sin36 = -2.746*10^6 N/C -ve y direction Magnitude = sqrt(Ex^2 + Ey^2) = 13.2 * 10^6 N/C For symmetry, Y component = 0. X component = 2Ecos 36 = 9.132 * 10^6 N/C Magnitude = 9.132*10^6 Force on electron = eE = 1.463 * 10^-12 N

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