Answer:
B
Explanation:
Given:-
- The charge of the test particle q = 3.0 * 10^-9 C
- The force exerted by the metal sphere F = 6.0 * 10^-5 N
Find:-
The magnitude and direction of the electric field
strength at this location?
Solution:-
- The relationship between the electrostatic force F exerted by the metal sphere on the test-charge and the Electric Field strength E at the position of test charge is given by:
F = E*q
- Using the data given we can determine E:
E = F / q
E = (6.0 * 10^-5) / (3.0 * 10^-9)
E = 20,000 N/C
- The direction of electric field is given by the net charge of the source ( metal sphere). The metal sphere is negative charge hence the direction of Electric Field strength E is directed towards the metal sphere.
<span>a)
Capacitance = k x ε° x area / separation
ε° = 8.854 10^-12 F/ m
k = 2.4max
average k = 0.78 / 1.27 * 2.4 +(1.27- 0.78) / 1.27 * 1 = 1.474 + 0.386 = 1.86
(61.4 % separation k = 2.4 --- 38.6 % k = 1 air --- average k = 0.614 * 2.34 + 0.386 * 1 = 1.86
area = 145 cm2 = 0.0145 m2
separation = 1.27 cm 0.0127 m
C = 1.86 * 8.854 10^-12 * 0.0145 / 0.0127 = 18.8 pF
b) Q = C * V --- 18.8 * 83 = 1560.4 pC = 1.5604 nC
c) E = V / d = 83 / 0.0127 = 6535.4 V/m </span>
The answer is,
<u>Farthest from the axis of rotation</u>
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It is helpful to study the light that comes off stars because C. The light from a star gives hints of what elements make up a star.
Answer: The field lines bend away from the second positive charge
Explanation: opposite attracts, same repulse