Calculate the electric potential at point B due to Q. 3.18 times 10^3 V 3.18 tim

Question

Calculate the electric potential at point B due to Q. 3.18 times 10^3 V 3.18 times 10^3 V 3.18 times 10^3 V 3.18 times 10^3 V None What is the magnitude of the electric force between the two charges? 4.50 times 10^6 N 4.50 times 10^9 V 4.50 times 10^3 N 4.50 times 10^8 N None A wire carries a 6.00-A current along the x-axis, and another wire carries a 8.00-A current along the y-axis, as shown in Fig-2. What is the net magnetic field at point P. located at x = 4.00 m. y = 3.00 m? In Fig-3 the battery emf is 50.0 V, the resistance is 250 ohm, and capacitance is 0.50 mu F. The switch S is closed for a long time interval, and zero difference is measured across the capacitor. After the switch is opened, the potential difference across the capacitor reaches a maximum values of 150 V. What is the inductance? An ac generator with rms voltage of 120-V and frequency of 60.0 Hz is connected in series with a 175.0-ohm resistor, a 15-mu F capacitor and a 50-mH inductor. Calculate the following parameters: (a) angular frequency of the circuit; (b) resonance frequency of the circuit (c) the impedance; and (d) the phase angle.

Explanation / Answer

8 )

V = kq/r1 + k*2q/r2

r1 = sqrt(2)*d

r2 = d

v = kq/d * ( 1/sqrt(2) + 2)

v = 2.436 * q/d

put the value of q and d

part 3)

w = 2pi*f

f = 60 HZ

w = 376.99 = 377 rad/s

part b )

at resonance

XL = Xc

w*L = 1/w*c

f = 1/2pi *sqrt(LC)

f = 183.78 Hz

part c )

Z = sqrt(R^2 + (XL - Xc)^2

Z = 235.76 ohm

part d )

phase angle = tan phi = (XL-Xc)/R

phi = -42 degree

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