You have a system of a positively and a negatively charged objects (+q and -9.) separated by distance d. (a) What is the sign of
the electric potential energy? (b) What can you do to increase the electric potential energy of the system? (C) What can you do to decrease the electric potential energy of the system? Support your answer with a bar chart and mathematically. Specify initial and final states on the bar chart.
1 answer:
Answer:
a)The sign of energy is negative, b) we can approach the two charges, c) d ⇒ 0
Explanation:
a) The elective potential energy is
U = k q₁q₂ / r
Applied to this case.
U = -k q 9 / d
U = -9 k q / d
The sign of energy is negative, which corresponds to a stable system
b) c) for the energy to be more negative we can approach the two charges
c) To increase the energetic of the system, that is to say less negative, we can separate the discharges whereby U approaches zero or we can decrease the value of the charge q.
d ⇒ 0
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Answer:
a) <em>2.278 x 10^-5 volts</em>
b) <em>1.139 x 10^-6 Ampere</em>
c) <em>2.59 x 10^-11 W</em>
Explanation:
The radius of the wire r = 2 mm = 0.002 m
the number of turns N = 200 turns
direction of the magnetic field ∅ = 25°
magnetic field strength B = 0.02 T
varying time = 2 sec
The cross sectional area of the wire =
==> A = 3.142 x = 1.257 x 10^-5 m^2
Field flux Φ = BA cos ∅ = 0.02 x 1.257 x 10^-5 x cos 25°
==> Φ = 2.278 x 10^-7 Wb
The induced EMF is given as
E = NdΦ/dt
where dΦ/dt = (2.278 x 10^-7)/2 = 1.139 x 10^-7
E = 200 x 1.139 x 10^-7 = <em>2.278 x 10^-5 volts</em>
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b) If the two ends are connected to a resistor of 20 Ω, the current through the resistor is given as
= E/R
where R is the resistor
= (2.278 x 10^-5)/20 = <em>1.139 x 10^-6 Ampere</em>
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<em> </em>c) power delivered to the resistor is given as
P = E
P = (1.139 x 10^-6) x (2.278 x 10^-5) = <em>2.59 x 10^-11 W</em>