Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

The capacitance of a variable radio capacitor can be changed from 50PF to 95PF by turning the dial from 0 to 180 degrees. With the dial set at 180 degrees , the capacitor is connected to a 400V battery. After charging the capacitor is disconnected from the battery and the dial is turned to 0 degrees. a - What is the charge on the capacitor b- What is the p.d across the capacitor when the dial reads 0 degrees. c - WHat is the energy of the capacitor in this position?

I got my questions answered at in under 10 minutes. Go to now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly


Get your free account and access expert answers to this and thousands of other questions

a. Q=CV=(95 10^-12)(400) = 38 10^-9 coulomb b.760 volts c. Use approriate equation (I don't have it memorized)
@radar how did you arrive at your answer pls explain @experimentX @RaphaelFilgueiras @Algebraic! any info
which part?

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

c) E=1/2 C V²
@woleraymond are you there?
yep im here
what you didnt understood?
how did @radar arrive at 760 volts for question b..also what impact does the turning have on the capacitor
the charge will be the same,when you turn the dial to zero degrees ,you will change the capacitance for 50 PF
then q=c.v-> 38E-9=50E-12.v->v=760
now energy will be 1/2.c.v²=1/2. 50E-12.(760²)
14.4 microjoules

Not the answer you are looking for?

Search for more explanations.

Ask your own question