Question
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1. A completely ionized beryllium atom (net charge = +4e) is accelerated through a potential difference of 6.0 V. What Is the increase in kinetic energy of the atom? 2. A charge q = -4.0 muC is moved 0.25 m horizontally to point P in a region where an electric field is 150 V/m and directed vertically as shown. What is the change in the electric potential energy of the charge? P and Q are points within a uniform electric field that are separated by a distance of 0.1 m as shown. The potential difference between P and Q is 50V. 3. Determine the magnitude of this electric field, 4. How much work is required to move a +1000muC Point charge from P to Q? 5. A charge is located at the center of sphere A (radius RA= 0.0010 m), which is in the center of sphere B (radius RB = 0.0012 m). Spheres A and B are both equipotential surfaces. What is the ratio VA/VB of the potentials of these surfaces?
Explanation / Answer
1. KE = eV
kE = 4 *1.67 e-19 * 6
KE = 24 eV ---> option e
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2. W = U = vq
but V = Ed
so W = Edq
W = 150 * 0.25 * 4 e-6
W = -1.5 * 10^-4 ---> option B
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3. V = Ed
E = 50/0.1
E = 500 V/m ---> option d
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4. W = Vq
W = 500* 1000 e-6
W = 0.05 J option B
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apply V = KQ/r
Va/Vb = rb/ra
Va/Vb = 0.0012/0.001
Va/ Vb = 1.2 option C
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