All categories

2 years ago

If the electric potential in a region is given by v(x)=7/x2 the x component of the electric field in that region is

Physics

1 answer:

lions [1.4K]2 years ago

6 0

Hi there!

Recall that:
$\Delta V = -\int\limits^a_b {E \cdot } \, dx$

Given an electric field, the potential difference can be solved by using integration. Similarly:
$E = -\frac{dV}{dx}$

We can differentiate the electric potential equation to solve for the electric field.

Use the power rule:
$\frac{dy}{dx} x^n = nx^{n - 1}$

Differentiate the given equation.

$-\frac{dV}{dx}\frac{7}{x^2} =- \frac{dV}{dx}7x^{-2} = -(-14x^{-3}) = \frac{14}{x^3}$

Or:
$\boxed{E(x) = \frac{14}{x^3}}$

You might be interested in

What does the negative sign in F = –kx mean?

Andru [333]

The force is opposite to the displacement

3 0

3 years ago

If a light is moved twice (2x) as far from a surface, the area the light covers is ___ as big.

Ksenya-84 [330]

Answer:

twice

Explanation:

From magnification = height of image / height of object

Distance of image/ distance of object = magnification

If the distance and height of the object represents the initial light distance and the exposed surface respectively.

And similarly the distance and height of the image represents the final light distance and the exposed surface respectively.

Hence the new image exposure would be twice as large.

If we use the formula our point of investigation is Height of image,

H2= D2/D1× H1

H2 = 2D2/D1 × H1

H2 = 2H1

6 0

3 years ago

Two parallel wires carry currents in the same direction. If the currents in the wires are 1A and 4A and the wires are 5 m apart.

serious [3.7K]

Answer:

$1.6\times 10^{-7}$ N

$2.4\times 10^{-7}$ N

Explanation:

$i_{1}$ = 1 A

$i_{2}$ = 4 A

$r$ = distance between the two wire = 5 m

$F$ = Force per unit length acting between the two wires

Force per unit length acting between the two wires is given as

$F = \frac{\mu _{o}}{4\pi }\frac{2i_{1}i_{2}}{r}$

$F = (10^{-7})\frac{2(1)(4)}{5}$

$F = 1.6\times 10^{-7}$ N

$r'}$ = distance of each wire from the midpoint = 2.5 m

Magnetic field midway between the two wires is given as

$B = \frac{\mu _{o}}{4\pi } \left \left ( \frac{2i_{2}}{r'} \right - \frac{2i_{1}}{r'} \right \right ))$

$B = (10^{-7}) \left \left ( \frac{2(4)}{2.5} \right - \frac{2(1)}{2.5} \right \right ))$

$B = 2.4\times 10^{-7}$

5 0

3 years ago

a ball of mass 100g moving at a velocity of 100m/s collides with another ball of mass 400g moving at 50m/s in same direction, if

klio [65]

Answer:

Velocity of the two balls after collision: $60\; \rm m \cdot s^{-1}$ .

$100\; \rm J$ of kinetic energy would be lost.

Explanation:

<h3>Velocity</h3>

Because the question asked about energy, convert all units to standard units to keep the calculation simple:

Mass of the first ball: $100\; \rm g = 0.1\; \rm kg$ .
Mass of the second ball: $400\; \rm g = 0.4 \; \rm kg$ .

The two balls stick to each other after the collision. In other words, this collision is a perfectly inelastic collision. Kinetic energy will not be conserved. The velocity of the two balls after the collision can only be found using the conservation of momentum.

Assume that the system of the two balls is isolated. Thus, the sum of the momentum of the two balls will stay the same before and after the collision.

The momentum of an object of mass $m$ and velocity $v$ is: $p = m \cdot v$ .

Momentum of the two balls before collision:

First ball: $p = m \cdot v = 0.1\; \rm kg \times 100\; \rm m \cdot s^{-1} = 10\; \rm kg \cdot m \cdot s^{-1}$ .
Second ball: $p = m \cdot v = 0.4\; \rm kg \times 50\; \rm m \cdot s^{-1} = 20\; \rm kg \cdot m \cdot s^{-1}$ .
Sum: $10\; \rm kg \cdot m \cdot s^{-1} + 20 \; \rm kg \cdot m \cdot s^{-1} = 30 \; \rm kg \cdot m \cdot s^{-1}$ given that the two balls are moving in the same direction.

Based on the assumptions, the sum of the momentum of the two balls after collision should also be $30\; \rm kg \cdot m \cdot s^{-1}$ . The mass of the two balls, combined, is $0.1\; \rm kg + 0.4\; \rm kg = 0.5\; \rm kg$ . Let the velocity of the two balls after the collision $v\; \rm m \cdot s^{-1}$ . (There's only one velocity because the collision had sticked the two balls to each other.)

Momentum after the collision from $p = m \cdot v$ : $(0.5\, v)\; \rm kg \cdot m \cdot s^{-1$ .
Momentum after the collision from the conservation of momentum: $30\; \rm kg \cdot m \cdot s^{-1}$ .

These two values are supposed to describe the same quantity: the sum of the momentum of the two balls after the collision. They should be equal to each other. That gives the equation about $v$ :

$0.5\, v = 30$ .

$v = 60$ .

In other words, the velocity of the two balls right after the collision should be $60\; \rm m \cdot s^{-1}$ .

<h3>Kinetic Energy</h3>

The kinetic energy of an object of mass $m$ and velocity $v$ is $\displaystyle \frac{1}{2}\, m \cdot v^{2}$ .

Kinetic energy before the collision:

First ball: $\displaystyle \frac{1}{2} \, m \cdot v^2 = \frac{1}{2}\times 0.1\; \rm kg \times \left(100\; \rm m \cdot s^{-1}\right)^{2} = 500\; \rm J$ .
Second ball: $\displaystyle \frac{1}{2} \, m \cdot v^2 = \frac{1}{2}\times 0.4\; \rm kg \times \left(50\; \rm m \cdot s^{-1}\right)^{2} = 500\; \rm J$ .
Sum: $500\; \rm J + 500\; \rm J = 1000\; \rm J$ .

The two balls stick to each other after the collision. Therefore, consider them as a single object when calculating the sum of their kinetic energies.