A particle of charge 2.0 x 10^-8C experiences an upward force of magnitude 4.0 x10^-6 when it is placed in a particular point in an electric field. a. What is the electric field at that point?
b. If a charge q=1.0 x 10^-8 C is placed there, what is the force on it?

(a) The electric field of the point charge is 200 N/C.

(b) When the charge changes, the force on the charge in the field is 2 x 10⁻⁶ N.

The electric field of the point charge in a given is defined as the force per unit of charge.

When the charge changes, the force on the charge in the field is calculated as follows;

F = Eq

F = (200) x (1 x 10⁻⁸)

F = 2 x 10⁻⁶ N

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a) The electric field at that point is newtons per coulomb.

b) The electric force is newtons.

Explanation:

a) Let suppose that electric field is uniform, then the following electric field can be applied:

(1)

Where:

- Electric field, measured in newtons per coulomb.

- Electric force, measured in newtons.

- Electric charge, measured in coulombs.

If we know that and , then the electric field at that point is:

The electric field at that point is newtons per coulomb.

b) If we know that and , then the electric force is:

The electric force is newtons.

real, smaller and upside-down.   so the answer should be b.

speed of light in vacuum is 300,000 km/s

speed of light in other mediums is given by the expression v = c/n, where c is the speed of light in vacuum and n is the refractive index of medium.

for the first medium v = 200,000 km/s

  so, refractive index of medium = 300000/200000 = 1.5

    1.5 refractive index is usually for glass. so first medium is glass.

for the second medium v = 250,00 km/s

so, refractive index of medium = 300000/250000 = 1.2

    1.2 refractive index is usually for water. so second medium is water.

for the third medium v = 0 km/s

so, refractive index of medium = 300000/0 = ∞

    ∞ refractive index is usually for opaque mediums. so third medium is opaque medium.