At a certain distance D from a stationary point charge Q, the magnitude of the f

Question

At a certain distance D from a stationary point charge Q, the magnitude of the field intensity is 400 V/m and the potential (w.r. to zero potential at infinite distance) is -2000 V. From the above information, find Q (magnitude and sign) and D. Show your reasoning. If an object of mass 10 mg carrying a negative charge 1.50 muC is released from rest at this point (where we have calculated the field and the potential), which way will it begin to move? Why? What will be the distance of the object (from the stationary charge Q) when it attains a speed of speed of 12.0 m/s Show your reasoning and calculations.

Explanation / Answer

(1). field intensity E = 400 V / m Potential V = -2000 V we know E = -V / r from this distance   r = -V / E = 5 m we know V = KQ / r from this charge Q = Vr / K where K = coulomb's constant = 8.99 * 10 ^9 Nm^ 2/ C^ 2 plug the values we get Q = 1.112*10^-6 C (2). mass m = 10 mg = 10 * 10 ^-3 g = 10* 10 ^-3 * 10 ^-3 kg = 10^-5 kg charge q = 1.5 *10^-6 C it moves opposite to the field direction . Since the force act on negative charge is opposite to the field direction. (3).Speed v = 12 m / s initial speed u = 0 from law of conservation of energy , Vq+ ( 1/ 2) mu^ 2 = V ' q + ( 1/ 2) mv^ 2                     V ' q = Vq - ( 1/ 2) mv^ 2                         V ' = V - ( 1/ 2q ) mv^ 2                               = -2000 -480                               = -2480 volt we know V ' = KQ / r' from this   r ' = KQ / V '                     = 4.031 m

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