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3 years ago

What is the equation for calculating the electrical force, Fe, between two charges?

1 answer:

4 0

When 2 electric charges are place close to each other they experience a force between them. This force may be repulsive or attractive depending on the type of charges involved.

The magnitude of this force depend on the quantity of individual charges and the distance between them.

The force is defined by the <em>Coulomb's law</em> states that: <em>The magnitude of the electrostatic force of attraction or repulsion between two point charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them.</em>

The formula for calculating the force between 2 electric charges is ;

F = (kq₁q₂)/d²

Where k is the proportionality constant known as Coulomb's constant,

q₁ and q₂ are the charges and

d is the distance between q₁ and q₂.

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Answer:

Explanation:

Since the block is at rest in an elevated position, we can assume that it only has potential energy.

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Answer:

The ball will fall back and land to Elle's hands.

Explanation:

The bus move in a straight line with constant velocity means that there is no change of direction and no acceleration. Inertia can change the direction of the ball and acceleration can change its velocity. Since these two factors is not present in this scenario, the ball only has vertical movement. Thus the ball will land where it was thrown, in Elle's hands.

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3 years ago

A particle with charge − 2.74 × 10 − 6 C −2.74×10−6 C is released at rest in a region of constant, uniform electric field. Assum

s2008m [1.1K]

Answer:

241.7 s

Explanation:

We are given that

Charge of particle= $q=-2.74\times 10^{-6} C$

Kinetic energy of particle= $K_E=6.65\times 10^{-10} J$

Initial time= $t_1=6.36 s$

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We have to find the time t after that the particle is released and traveled through a potential difference 0.351 V.

We know that

$qV=K.E$

Using the formula

$2.74\times 10^{-6}V_1=6.65\times 10^{-10} J$

$V_1=\frac{6.65\times 10^{-10}}{2.74\times 10^{-6}}=2.43\times 10^{-4} V$

Initial voltage= $V_1=2.43\times 10^{-4} V$

$\frac{\initial\;voltage}{final\;voltage}=(\frac{initial\;time}{final\;time})^2$

Using the formula

$\frac{V_1}{V_2}=(\frac{6.36}{t})^2$

$\frac{2.43\times 10^{-4}}{0.351}=\frac{(6.36)^2}{t^2}$

$t^2=\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}$

$t=\sqrt{\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}}$

$t=241.7 s$

Hence, after 241.7 s the particle is released has it traveled through a potential difference of 0.351 V.

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