nautica
1) A current of .5 A exist on a 60-ohm lamp. Find Potential Difference

2) The sum of currents into a junction equals the sum of the currents out of the junction.

D)convservation of energy
E) Conservation of charge My answer is E

3)Current is a measure of:
D) amount of charge that moves past a point per unit time
E) speed with which a charge moves past a point

4) A 2uF and a 1uF capacitor are connected in a series and a potential difference is applied across the combination. The 2uF capacitor has:
a) twice the charge of the 1uF
b) half the charge of the 1uF
c) twice the potential difference of the 1uF
d) half the potential difference of the 1uF
e) none of the above

C=q/v so I picked A

5) the units of the electric field are
a) J/(C*m)
b) J/C
c) J*C
d) J/m
e) none of these

I know that electric field is N/C, But, I am think that would be the same as J/(C*m) So I say if not A then E

6) The electric field at a distance of 10 cm from an isolated point charge of (2x10^9 C) is: I cant even find a formula for this one.

7) A car battery is rated at 80 A * h is a unit of:
a) power
b) energy
c) current
d) charge
e) force

My ansswer is A) power

8) The capacitance of a parrallel-plate capacitor can be increase by:

A) increasing the charge
b) decreasing the charge
c) increasing the plate separation
d) decreasing the plate separation
e) decreasing the plate area

C=E A/d So I assume that if you decrease distance you will in crease capacitance. So d) decreast the plate separation.

Thanks
Nautica

Related Introductory Physics Homework Help News on Phys.org
For Pro 4: Charges will remain constant as capacitors are connected in series (**Look & Solve it Again**)

For Pro 5u have

energy =1/2 CV2
also Q=CV=CEd from above to eqn it seems A is correct if C is used for Capacitance

For Pro 6
electric field dur to a point charge is

$$=\frac{Q}{4\pi \epsilon_0 R^2}$$

radially Outwards for + charge and inwards for - charge

Rest are correct

chroot
Staff Emeritus
Gold Member
Number 7 is wrong. The amp-hour is a unit of charge. When multiplied by the potential created by the battery, you get energy.

- Warren

nautica
Thanks

4) A 2uF and a 1uF capacitor are connected in a series and a potential difference is applied across the combination. The 2uF capacitor has:
a) twice the charge of the 1uF
b) half the charge of the 1uF
c) twice the potential difference of the 1uF
d) half the potential difference of the 1uF
e) none of the above

C=q/v so if "charges remain constant", then then answer is D

5) the units of the electric field are
a) J/(C*m) Joules/(Coulomb times meters)
b) J/C
c) J*C
d) J/m
e) none of these

The C is for Coulomb not capacitance, so is electric feild is N/C then I can divide by m which convert the Newton to a Joule and multiply the coulomb by a meter, which would be A - right?

6) The electric field at a distance of 10 cm from an isolated point charge of (2x10^9 C) is: I cant even find a formula for this one.

The Q/(4pieER^2)is what was think but there is no value for E. The choices are as follows:

a) 1.8 N/C
b) 180 N/C
c) 18 N/C
d) 1800 N/C
e) none of the above.

7) A car battery is rated at 80 A * h is a unit of:
a) power
b) energy
c) current
d) charge
e) force

Okay, you said the answer for this one is energy. I am still a little confused. Coulomb is a Force. and Amp is 1 coulomb/sec to if we multply this by time (s) then it would be Force/s^2. which to me would be similar to a Watt, which is a measure of Power.????

Once again, thanks for the help
Nautica

chroot
Staff Emeritus
Gold Member
Originally posted by nautica
The C is for Coulomb not capacitance... which would be A - right?
Right.
The Q/(4pieER^2)is what was think but there is no value for E. The choices are as follows:
No values for E? The electric field, E, is measured in units of N/C. Hint: I think the question asked for 2*10^-9 C, NOT 2*10^9 C.
7) A car battery is rated at 80... Okay, you said the answer for this one is energy.
No, I quite clearly said the ampere-hour is a unit of charge.
I am still a little confused. Coulomb is a Force.
No, it's not. The coulomb is a unit of charge.

- Warren

Q/(4pieER^2)
Why there is E in the denominator of the equation.

An Electric force is given by coulombs Law not by Coulomb(Unit of Charge)