1) Find the magnitude and direction of the electric field from each of the three

Question

1) Find the magnitude and direction of the electric field from each of the three charges at point A.

2) Find the magnitude of the total electric field at point A.

3) Find the direction of the total electric field at point A makes with the horizontal.

4) Find the electric potential at point A (relative to ?) caused by the three charges.

5) If a charge q = +70 ?C were located at point A, find the electric force on the q = +70 ?C charge.

6) If a charge q = +70 ?C charge has a mass m = 17.0 ?C and starts fro rest at point A, what will its velocity be very far from all three charges (at ?).

*** I appreciate your help big time here! I am having a very hard time understanding these types of questions so I want to thank you in advance for helping! ***

Explanation / Answer

Part 1)

From the 80uC charge on the left

E = kq/r2 = (9 X 109)(80 X 10-6)/(.025)2

E = 1.152 X 109 N/C to the right

From the -20uC charge on the right

E = kq/r2 = (9 X 109)(20 X 10-6)/(.025)2

E = 2.88 X 108 N/C to the right

From the -40uC charge at the top

E = kq/r2 = (9 X 109)(40 X 10-6)/(.15)2

E = 1.6 X 107 N/C upward

Part 2)

The net field to the right is 2.88 X 108 + 1.152 X 109 = 1.44 X 109 N/C

Put that into the pyhtagorean theorem with the downward Field and get

Net2 = (1.44 X 109)2 + (1.6 X 107)2

Net = 1.44 X 109 N/C

Part 3)

The direction is from the tangent function

tan(angle) = 1.6 X 107/1.44 X 109

Angle = 0.637o Below the Horizontal

Part 4)

V = kq/r for all three charges

V = kq/r + kq/r + kq/r

V = (9 X 109)(80 X 10-6/.025) + (-20 X 10-6)/.025 + (-40 X 10-6)/.15

V = 1.92 X 107 V

Part 5)

F = qE

F = (70 X 10-6)(1.44 X 109)

F = 100800 N

Part 6)

Since the mass makes no sense in part 6, here is how you set it up

PE = KE

qV = .5mv2

(70 X 10-6)(1.92 X 107) = (.5)(m)(v2)

2688 = mv2

v = sqrt (2688/(17 X 10-9)

v = 3.98 X 105 m/s

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