Two positive point charges are placed as shown in diagram. At point P the magnit

Question

Two positive point charges are placed as shown in diagram. At point P the magnitude of the electric field due to charge A is 6.0 times 10^6 N/C. The magnitude of the electric field due to charge B at point P: 4.0 times 10^6 N/C. Determine the net electric field at point P. Determine the maximum speed an alpha particle could reach if it moves from rest through a potential difference of 4.20 kV. A 1.8 times 10^-9 kg charged oil drop accelerates upwards in a vertical electric field of strength 500 N/C. When its acceleration is upwards at 3.0 m/s^2 the charge on the oil drop must be? Determine the maximum speed an alpha particle could reach if it moves from rest through a potential difference of 4.20 kV.

Explanation / Answer

a) Net electric field at P, Enet = [(6 * 106)2 + (4 * 106)2]1/2 = 7.2 * 106 N/C

b) Max. KE of alpha particle = qV

=> mvmax2/2 = (2e)V

=> vmax = (4eV/m)1/2 = [4 * (1.6 * 10-19) * (4.20 * 103) / (6.64 * 10-27)]1/2 = 6.36 * 105 m/s

c) Acceleration, a = qE/m

=> q = ma/E = (1.8 * 10-9) * 3 / 600 = 9 * 10-12 C

d) Same question as part b.

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