Respuesta :
The equation for capacitors to look for the charge is Q=CV.
We are given 10 μF and 1.7 kV, so we know that C = 10 x 10^-6 F and V=1.7x10^3 V.
As an outcome, we can multiply those two numbers to get Q = .017C
Q = CV
Q = (10 x 10^-6 F) (1.7 x 10^3 V)
Q = .017 C
The amount of charge stored in the capacitor is [tex]\boxed{0.017\,{\text{C}}}[/tex].
Further Explanation:
Given:
The capacitance of the capacitor is [tex]10\mu {\text{f}}[/tex] .
The voltage applied to the capacitors by the laser beamis [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 charge stored in each of the capacitor will also remain same.
The amount of charge stored in a capacitor is given by:
[tex]Q = CV[/tex]
Here, [tex]Q[/tex] is the amount of charge 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}Q&= \left( {10 \times {{10}^{ - 6}}\,\mu {\text{f}}\right)\times{\tex\left( {1.7 \times{{10}^3}\,{\text{V}}}\right)\\&= 0.017\,{\text{C}}\\\end{aligned}[/tex]
Therefore, the amount of charge stored in the capacitor is [tex]\boxed{0.017\,{\text{C}}}[/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, charge store, capacitance, 0.017 J, Q=CV.
