A 7uF capacitor is charged to a potential difference of 110V The electrical energy stoed in the capacitor is
(4.2x10^-2 was the given answer)
but I used C=Q/V and keep getting 7.7x10^-4
also A particle of charge Q and mass m is accelerated from rest through a potential difference V, attaining kinetic energy K. What is the kinetic energy of a particle of charge 2Q and mass m/2 that is accelerated under the same conditions? (Answer was given as 2K but i got 4k)
W = 0.5CV^2,
W=0.5 * 7*10^-6 * (110)^2 = 4.23*10^-2.
To find the electrical energy stored in the capacitor, you need to use the formula:
Electrical Energy = (1/2) * C * (V^2)
Given that the capacitance (C) is 7uF and the potential difference (V) is 110V, we can substitute these values into the formula:
Electrical Energy = (1/2) * 7uF * (110V)^2
Now, let's convert the capacitance to farads and simplify the equation:
Electrical Energy = (1/2) * 7 * 10^-6F * (110V)^2
= (1/2) * 7 * 10^-6 * 12100
= 7 * 10^-6 * 6050
= 42.35 * 10^-3
= 4.235 * 10^-2 J
Therefore, the correct answer is 4.235x10^-2 J, which is approximately equal to 4.2x10^-2 J.
Regarding the second question, let's compare the equations for kinetic energy:
For a particle of charge Q and mass m:
K = (1/2) * Q * V^2
For a particle of charge 2Q and mass m/2:
K' = (1/2) * (2Q) * V^2
Now, let's simplify K' to determine the expression for the kinetic energy:
K' = (1/2) * 2Q * V^2
= Q * V^2
Comparing K and K', we can see that K' = K. Therefore, the kinetic energy of a particle with charge 2Q and mass m/2 under the same conditions is equal to 2K, not 4K as you initially calculated.
Hence, the correct answer is 2K.