Respuesta :
Energy stored= 1/2 * capacitance * voltage^2
= 1/2 * 10*10^-10* (1.7*1000)^2
= 1.7*1.7/2000 J
= 1/2 * 10*10^-10* (1.7*1000)^2
= 1.7*1.7/2000 J
The amount of energy stored in the capacitor is [tex]\boxed{14.45\,{\text{J}}}[/tex].
Further Explanation:
Given:
The capacitance of the capacitor is [tex]10\,{\text{\mu f}}[/tex].
The voltage applied to the capacitors by the laser beam is [tex]1.7\,{\text{kV}}[/tex].
Concept:
Since the pair of capacitor is identical and the voltage applied across the capacitors is same then the amount of energy associated with each of the capacitor will also remain same.
The amount of energy stored in a capacitor is given by:
[tex]E = \dfrac{1}{2}C{V^2}[/tex]
Here, [tex]E[/tex] is the amount of energy stored in the capacitor, [tex]C[/tex] is the capacitance of the capacitor and [tex]V[/tex] is the voltage drop across the capacitor.
Substitute the values of capacitance and the voltage supplied across the capacitor in the above expression:
[tex]\begin{aligned}E&=\frac{1}{2}\times \left(10\times 10^{-6}\mu\text{f}\right)\times\left(1.7\times 10^{3}\text{V}\right)^{2}\\&=\frac{28.9}{2}\\&=14.45\text{J}\end{aligned}[/tex]
Therefore, the amount of energy stored in the capacitor is [tex]\boxed{14.45\,{\text{J}}}[/tex].
Learn More:
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Answer Details:
Grade:College
Subject: Physics
Chapter:Capacitors
Keywords: Energy, stored, capacitor, pair of capacitors, 10uf, high power laser, 1.7kV, each capacitor, capacitance, 14.45 J, 1/2CV^2.
