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What is the displacement current in the capacitor if the potential difference across the capacitor is increasing at 500,000V/s?

konstantin123 [22]

Answer:

$I=2.71\times 10^{-5}\ A$

Explanation:

A 6.0-cm-diameter parallel-plate capacitor has a 0.46 mm gap.

What is the displacement current in the capacitor if the potential difference across the capacitor is increasing at 500,000V/s?

Let given is,

The diameter of a parallel plate capacitor is 6 cm or 0.06 m

Separation between plates, d = 0.046 mm

The potential difference across the capacitor is increasing at 500,000 V/s

We need to find the displacement current in the capacitor. Capacitance for parallel plate capacitor is given by :

$C=\dfrac{A\epsilon_o}{d}\\\\C=\dfrac{\pi r^2\epsilon_o}{d}$ , r is radius

Let I is the displacement current. It is given by :

$I=C\dfrac{dV}{dt}$

Here, $\dfrac{dV}{dt}$ is rate of increasing potential difference

So

$I=\dfrac{\pi r^2\epsilon_o}{d}\times \dfrac{dV}{dt}\\\\I=\dfrac{\pi (0.03)^2\times 8.85\times 10^{-12}}{0.46\times 10^{-3}}\times 500000\\\\I=2.71\times 10^{-5}\ A$

So, the value of displacement current is $2.71\times 10^{-5}\ A$ .

4 0

3 years ago

A ball of mass m is thrown into the air in a 45° direction of the horizon, after 3 seconds the ball is seen in a direction 30° f

Rzqust [24]

Answer:

Velocity (magnitude) is 98.37 m/s

Explanation:

We use the vertical component of the initial velocity, which is:

$v_{0y}=v_0*sin(45)=\frac{\sqrt{2} }{2}v_0$

Using kinematics expression of vertical velocity (in y direction) for an accelerated motion (constant acceleration, which is gravity):

$v_{y}=v_{0y}+a*t=\frac{\sqrt{2} }{2}v_0-9.8t$

Now we need to find $v_y$ as a function of $v_0$ . We use the horizontal velocity, which is always the same as follow:

$v_x=v_0cos(45\º)=\frac{\sqrt{2} }{2}v_0=v_{t=3}*cos(30\º) \\$

We know the angle at 3 seconds:

$v_y(t=3)=v_{t=3}*sin(30\º)\\v_{t=3}=\frac{v_y}{sin(30\º)}$

Substitute $v_{t=3}$ in $v_x$ and then solve for $v_y$

$\frac{\sqrt{2} }{2}v_0=\frac{v_y*cos(30\º) }{sin(30\º)} \\v_y=\frac{\sqrt{6} }{6}v_0$

With this expression we go back to the kinematic equation and solve it for initial speed

$\frac{\sqrt{6} }{6} v_0 =\frac{\sqrt{2} }{2}v_0-29.4\\v_0(\frac{\sqrt{6}-3\sqrt{2}}{6} )=-29.4\\v_0=98.37 m/s$

3 0

4 years ago

The variables for this experiment include mass, volume, and the materials in the various balls and their densities. In Part III,

zmey [24]

Answer:

volume and density

Explanation:

I took the test

set as brainliest

8 0

2 years ago

Read 2 more answers

The edge of a flying disc with a radius of 0.13 m spins with a tangential speed of 3.3 m/s.

Aleks [24]

By definition, centripetal acceleration is given by:

$a = \frac{v ^ 2}{r}$

Where,

v: tangential disk speed

r: disk radius

Substituting values in the given equation we have:

$a =\frac{3.3^2}{0.13}\\a = 83.76923077$

Rounding the result we have:

$a = 83.8 \frac{m}{s^2}$

Answer:

The centripetal acceleration of the disc edge in m/s^2 is:

$a = 83.8 \frac{m}{s^2}$

3 0

4 years ago

Read 2 more answers

How to find initial velocity without time?

77julia77 [94]

1. The problem statement, all variables and given/known data Knowing that snow is discharged at an angle of 40 degrees, determine the initial speed, v0 of the snow at A. Answer: 6.98 m/s 2. Relevant equations 3. The attempt at a solution I have found the x and y velocity and position formulas. Now since I don't know time, should I solve both position equations for time (t) and set them equal to each other to get my only unknown, vi? The quadratic equation for time in the y-dir seems a bit hectic. Is there an easier way to go about trying to find vi?

6 0

4 years ago