Three charged marbles are glued to a nonconducting surface and are placed in the
ID: 1405488 • Letter: T
Question
Three charged marbles are glued to a nonconducting surface and are placed in the diagram as shown. The charges of each marble are q1 = 5.90 µC, q2 = 1.42 µC,and q3 = 1.63 µC.Marble q1 is a distance r1 = 3.00 cm to the left of the marble q2, while marble q3 is a distance r3 = 2.00 cm to the right of the marble q2, as shown. Calculate the magnitude of the electric field a distance r' = 1.00 cm to the left of the center marble.
I got 2.44 e 8 N/C but this is wrong---i was given this hint
What is the electric field due to each individual marble? How do you find the total or "net" electric field when you have several electric fields?
Another marble is placed 1 cm to the left of the middle marble. If this new marble has a charge of 3.55 µC, calculate magnitude and direction of the force on it.
magnitude ?????? N direction is to the leftExplanation / Answer
(a)Firsrt of all we have to calcualte electric field at required point i.e observation point due to all charges
E1 = kQ1/r2 =( 9*109*5.9*10-6)/(0.02)2 = 132.75*106 N/C
E2 = kQ2/r22 = ( 9*109*1.42*10-6)/(0.01)2 = 127.8*106 N/C
E3 = kQ3/r32 = ( 9*109*1.63*10-6)/(0.03)2 = 163*106 N/C
Now we have to check the direction of all the electric field
Since charge 1 and 2 are positive therefore
charge 1 , electric field direction would be toeard + X
and charge 2 electric field direction would be toward -X direction.
Charge 3 is negative therefore its direction will be toward + X direction
Now Net electric field (ENET)= E1 + E3 - E2 = (132.75+163 - 127.8)*106 = 167.95*106 N/C
Since Electric field is a vector quantity therefore we have to add them vectorially.
If we have put the charge on the observation point then the force will be
F = qENET = 3.35*167.95*106 = 562.635*106 N/C
Since the electric field was toward + X direction and the charge is negative therefore force will be toward -X direction.
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