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10 months ago

In the diagram, q1 = +6.39*10^-9 C and

Physics

1 answer:

CaHeK987 [17]10 months ago

6 0

E =+823.12N/C is the electric field at point P

Each point in space has an electric field associated with it when a charge of any kind is present. The value of E, often known as the electric field strength, electric field intensity, or just the electric field, expresses the strength and direction of the electric field. in the diagram, q1 = +6.39*10^-9 C and

q2 = +3.22*10^-9 C hence E = 823.12 N/C

A region of space surrounding andd P Include a + or - sign toindicate the direction.P910.424 m-0.636 m- electrically charged particle or object known as an electric field is one in which an electric charge would experience force. A vector quantity called an electric field can be represented by arrows pointing in the direction of or away from charges. The force per unit charge exerted on a positive test charge that is at rest at a given position is the force per unit charge that is used to define the electric field analytically.

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A particle (charge = +0.8 mC) moving in a region where only electric forces act on it has a kinetic energy of 6.7 J at point A.

Maksim231197 [3]

Answer:

The kinetic energy of the particle as it moves through point B is 7.9 J.

Explanation:

The kinetic energy of the particle is:

$\Delta K = \Delta E_{p} = q\Delta V$

Where:

K: is the kinetic energy

$E_{p}$ : is the potential energy

q: is the particle's charge = 0.8 mC

ΔV: is the electric potential = 1.5 kV

$\Delta K = q \Delta V= 0.8 \cdot 10^{-3} C*1.5 \cdot 10^{3} V = 1.2 J$

Now, the kinetic energy of the particle as it moves through point B is:

$\Delta K = K_{f} - K_{i}$

$K_{f} = \Delta K + K_{i} = 1.2 J + 6.7 J = 7.9 J$

Therefore, the kinetic energy of the particle as it moves through point B is 7.9 J.

I hope it helps you!

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2 years ago

1. A 1.5 kg ball moves in a circle that is 0.5 m in radius at a speed of 5.1 m/s.

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Explanation: Solution attached:

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2 years ago

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