Question

Need help with capacitance:

Hey guys , so I've been reviewing the question and I really got stuck at some of the questions , I'll post them with my solution and you guys please help me if I was wrong:

*Note some of the question I could not solve them..

Answer 1

Q1/An isolated charged conducting sphere of radius 12 cm

creates an electric field of 4.90 x 10^4 N/C at a distance 21 cm

from its center. (a) What is its surface charge density? (b) What

is its capacitance?
a)
Surface q density = q/a and we want q so we use e=kq/r^2
Q=9*9^10/4.90*10^4*(21*10^-3)^2
after finding q we find the surface q denisity which is:
=Q/4*(3.14*(12*10^-3))^2

is it right?

b)
c=q/delta v

we find delta v which is
v=ed which is 4.90*10^4*(21*10^-3)
then we use it to find c.

is my solutions right ? I'll post the other questions..

Answer 2

Q2/An air-filled capacitor consists of two parallel plates, each

with an area of 7.6 cm2, separated by a distance of 1.80 mm. A

20 V potential difference is applied to these plates. Calculate (a)

the electric field between the plates, (b) the surface charge

density, (c) the capacitance, and (d) the charge on each plate.

a)
we use v=ed
so e=v/d
so e=20/0.0018000 we will get e.

b)
it's = to q/a
and we find q by:
q=k/e*r^2 like we did in question 1 and then we use it to find the surface charge which is
q/a so our q/0.00076m^2

c)
c=q/delva v we use our q then divide it by 20

d)
I don't know how to solve this :(

Answer 3

Q3/When a potential difference of 150 V is applied to the plates of a

parallel-plate capacitor, the plates carry a surface charge density of

30 nC/cm2. What is the spacing between the plates?

I don't know how to solve this really :(


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Q4/ Two capacitors, C1= 5 µ F and C2 =12 µ F, are connected in

parallel, and the resulting combination is connected to a 9 V battery.

(a) What is the equivalent capacitance of the combination? What

are (b) the potential difference across each capacitor and (c) the

charge stored on each capacitor?

a)
it's easy no need..

b)
how could I find the potential difference here ?
v=ed ??

c)
it's the same as the second question that I couldn't solve :(

Answer 4

Q1/An isolated charged conducting sphere of radius 12 cm

creates an electric field of 4.90 x 10^4 N/C at a distance 21 cm

from its center. (a) What is its surface charge density? (b) What

is its capacitance?
a)
Surface q density = q/a and we want q so we use e=kq/r^2
Q=9*9^10/4.90*10^4*(21*10^-3)^2
after finding q we find the surface q denisity which is:
=Q/4*(3.14*(12*10^-3))^2

is it right?

Answer 5

e=kq/r^2

\(q = er^2/k\)

\(e = 4.90 x 10^4 \)
\(k = 9*10^9\)
\(r = 21*10^-2\)

Answer 6

oh , thanks , I got it wrong :(

Answer 7

np :)

Answer 8

u want me check others ?

Answer 9

Yeah please :(

Answer 10

okay...
100cm = 1m
so, 1cm = 10^-2 m ok

Answer 11

it seems u were using 1cm = 10^-3 m in all ur calculations

Answer 12

oh boy :(

it's ok do not worry about that , the only thing I'm worried about is solving the problem itself , is my solution ok ?

I really hate this subject and thank god it's the last :)

Answer 13

Q2/An air-filled capacitor consists of two parallel plates, each

with an area of 7.6 cm2, separated by a distance of 1.80 mm. A

20 V potential difference is applied to these plates. Calculate (a)

the electric field between the plates, (b) the surface charge

density, (c) the capacitance, and (d) the charge on each plate.

Answer 14

d)
I don't know how to solve this :(

Answer 15

Multiply below two quantities :-
1) Area of plate
2) Answer from part b

Answer 16

why did you jump to d ? is my solution for the others correct ? thanks buddy

Answer 17

it looks okay, \(\epsilon_r = 1\) for air right ?

Answer 18

I don't really know as I said I hate this subject I'm not that good when it comes to this subject :(

Answer 19

its okay, you're just doing the dry electric field stuff...
the interesting magnetic fields/motors/generators starts soon, then u wil start to like the subject more

Answer 20

I'm afraid this is the last time I'll study this subject , since I'm majoring in computer science :)

anyways what is the solution for d ? and how about the other questions ?

Answer 21

you can work d) in atleast two ways :-

\(Q = CV\)

Answer 22

the other way, i have told u already

Answer 23

\(Q = CV\)

V = 20
C = Answer from part c

Answer 24

ah thanks , how about the other questions ?

Answer 25

I just wanted to say that I appreciate your effort to solve this , and helping me understand , thanks a lot :)

Answer 26

3rd q :-

Answer 27

oh thank you so so so much , you helped a lot :$