Two point charges, +2.80 μC and -6.20 μC, are separated by 2.30 m. What is the electric potential midway between them?

Physics

1 answer:

Rus_ich [418]4 years ago

6 0

Answer:

V=-26651.7V

Explanation:

The electric potential is for an electric charge is expressed as

V=kq/r.

Where K is constant, q is the charge and r is the distance.

Hence for a charge of +2.80 μC and - 6.20μC, both at distance 2.3m to each other, we can calculate the electric potential midway between them as

V=(kq/r)-(kq/r)

Since the mid way distance is 2.3/2=1.15m

Hence

V=K[(q/1.15) - (q/1.15)]

V=9*10⁹*10⁻⁶(2.43-5.39)

V=-26651.7V

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A person exerts a 19-N force on a cart attached to a spring and holds the cart steady. The cart is displaced 0.060 m from its eq

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To solve this problem, apply the concepts related to Hooke's law. From there we will find the spring constant. Subsequently, applying Energy balance, which includes gravitational potential energy, elastic potential energy and kinetic energy, we will bury the system's energy. Finally, using the displacement expression for the simple harmonic movement, we will find the expression that describes the system.

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