<span>Place a test charge in the middle. It is 2cm away from each charge.
The electric field E= F/Q where F is the force at the point and Q is the charge causing the force in this point.
The test charge will have zero net force on it. The left 30uC charge will push it to the right and the right 30uC charge will push it to the left. The left and right force will equal each other and cancel each other out.
THIS IS A TRICK QUESTION.
THe electric field exactly midway between them = 0/Q = 0.
But if the point moves even slightly you need the following formula
F= (1/4Piε)(Q1Q2/D^2)
Assume your test charge is positive and make sure you remember two positive charges repel, two unlike charges attract. Draw the forces on the test charge out as vectors and find the magnetude of the force, then divide by the total charge to to find the electric field strength:)</span>
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His is a step down transformer since n(primary) is greater than n(seconcary). You relate the input voltage with the ouput voltage with the following equation:
<span>Vout = n2/n1*Vin (n2/n1 is essentially your 'transfer function' that dictates what a specified input would produce) </span>
<span>Solving the equation: </span>
<span>Vin = Vout*n1/n2 = (320V)*(600/300) = 640 V </span>
<span>This is checked by seeing if Vin is greater than Vout, which it is for a step down transformer.</span>
Answer: F = 130 N
Explanation: Solution:
Convert first 27 cm to m.
27 cm x 0.01 m / 1 cm = 0.27 m
Calculate the torque using T = Fd
Derive to find force F
F = T /d
= 35 N.m / 0.27 m
= 130 N
