All categories

3 years ago

Calculate the electric potential energy in a capacitor that stores 4.0 x 10-10 C

Physics

2 answers:

serious [3.7K]3 years ago

5 0

5.0•10^-8J

Ape.x .........-_-

ioda3 years ago

4 0

The electric potential energy is $1\cdot 10^{-7} J$

Explanation:

The electric potential energy stored in a capacitor is given by the equation

$PE=Q \Delta V$

where

Q is the charge stored in the capacitor

$\Delta V$ is the potential difference between the plates

For the capacitor in this problem, we have

$Q=4.0\cdot 10^{-10}C$ is the charge

$\Delta V=250.0 V$ is the potential difference

Substituting, we find the potential energy:

$PE=(4.0\cdot 10^{-10})(250.0)=1\cdot 10^{-7} J$

Learn more about capacitors:

brainly.com/question/10427437

brainly.com/question/8892837

brainly.com/question/9617400

#LearnwithBrainly

You might be interested in

a 90 kg architect is standing 2 meters from the center of a scaffold help up by a rope on both sides. the scaffold is 6m long an

Mademuasel [1]

We can solve the problem by requiring the equilibrium of the forces and the equilibrium of torques.

1) Equilibrium of forces:
$T_1 - W_p - W_s + T_2 =0$
where
$W_p = (90kg)(9.81 m/s^2)=883 N$ is the weight of the person
$W_s = (200kg)(9.81 m/s^2)=1962 N$ is the weight of the scaffold
Re-arranging, we can write the equation as
$T_1 = 2845 N-T_2$ (1)

2) Equilibrium of torques:
$T_1 \cdot 3 m - W_p \cdot 2 m - T_2 \cdot 3m =0$
where 3 m and 2 m are the distances of the forces from the center of mass of the scaffold.
Using $W_p = 883 N$ and replacing T1 with (1), we find
$2845 N \cdot 3 m - T_2 \cdot 3 m - 833 N \cdot 2 m - T_2 \cdot 3 m=0$
from which we find
$T_2 = 1128 N$

And then, substituting T2 into (1), we find
$T_1 = 1717 N$

8 0

3 years ago

A very long solid insulating cylinder has radius R = 0.1 m and uniform charge density rho0= 10-3 C/m3. Find the electric field a

Galina-37 [17]

Answer:

$E = (0.56 \times 10^8 ) r \ \ N/c$

Explanation:

Given that:

$\rho_o = (10^{-3} ) \ c/m^3$

R = (0.1) m

To find the electric field for r < R by using Gauss Law

${\oint}E^{\to}* da^{\to} = \dfrac{Q_{enclosed}}{\varepsilon_o} --- (1)$

For r < R

$Q_{enclosed}=(\rho) ( \pi r^2 ) l$

$E*(2 \pi rl)= \dfrac{\rho ( \pi r ^2 l)}{\varepsilon_o}$

$E= \dfrac{\rho ( r)}{2 \varepsilon_o}$

where;

$\varepsilon_o = 8.85 \times 10^{-12}$

$E= \dfrac{10^{-3} ( r)}{2 (8.85 \times 10^{-12})}$

$E = (0.56 \times 10^8 ) r \ \ N/c$

4 0

3 years ago

If a ball with an original velocity of zero is dropped from a tall structure and it takes 7 seconds to hit the ground, what velo

Colt1911 [192]

The velocity of the ball when it reaches the ground is equal to B. 68.6 m/s. This value was obtained from the formula Vf = Vi + at. Vf is the final velocity. Vi is the initial velocity. The acceleration is "a", while the time of travel is "t". The solution is:

<span>Vf = Vi + at
</span>Vf = 0 + (-9.8 m/s^2) (7 s)
Vf = -68.6 m/s

The negative sign denotes the direction of the ball.

5 0

3 years ago

Read 2 more answers

How does beam spreading affect the temperature on earth and the seasons

navik [9.2K]

Get to school and learn boi

8 0

3 years ago

Show your working please

saw5 [17]

Explanation:

There's not enough information in the problem to solve it. We need to know either the initial speed of the lorry, or the time it takes to stop.

For example, if we assume the initial speed of the lorry is 25 m/s, then we can find the rate of deceleration:

v² = v₀² + 2aΔx

(0 m/s)² = (25 m/s)² + 2a (50 m)

a = -6.25 m/s²

We can then use Newton's second law to find the force:

F = ma

F = (7520 kg) (-6.25 m/s²)

F = -47000 N

3 0

3 years ago