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dangina [55]
3 years ago
9

In a Millikan oil-drop experiment, a uniform electric field of 5.71 x 10^5 N/C is maintained in the region between two plates se

parated by 6.49 cm. Find the potential difference (in V) between the plates.
Physics
2 answers:
faust18 [17]3 years ago
4 0

Answer:

V = 3.71×10⁴ V

Explanation:

Potential difference: This can be defined as the work done in moving a positive charge from infinity to any point in an electric field.

The S.I unit of potential difference is Volt (V).

The expression for potential difference is

V = E×d.............................. Equation 1

Where V = potential difference between the plates, E = Electric field , d = distance of separation between the plates

Given: E = 5.71×10⁵ N/C, d = 6.49 cm = 0.0649 m.

Substitute into equation 1

V = 5.71×10⁵×0.0649

V = 3.71×10⁴ V

Pani-rosa [81]3 years ago
3 0

Answer:

37057.9V

Explanation:

Electric potential is defined as the work done in moving a unit positive charge from infinity to a point.

Electric potential (E) = Potential Difference (V)/distance between plates(d)

Given; electric field of 5.71 x 10^5 N/C; distance between plates =6.49cm = 0.0649m

Since E = V/d

V = Ed

V = 5.71×10^5×0.0649

V = 37057.9Volts

The potential difference (in V) between the plates is 37057.9V

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Answer:

a. B= 9.45 \times10^{-3} T

b. B= 0.820 T

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Explanation:

First, look at the picture to understand the problem before to solve it.

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μ0 is the constant of proportionality

μ0=4πX10^-7 N*s2/c^2

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Here, the point is located in the surface of the cilinder. Therefore, we have to consider the current density of the conductor to calculate the magnetic field as follows:

J: current density

c: outer radius

b: inner radius

The cilinder's current is negative, as it goes on opposite direction than the wire's current.

J= \frac {-I_{c}}{\pi(c^{2}-b^{2}  ) }}

J=\frac{-36}{\pi(5.33\times10^{-5}-1.16\times10^{-5}) } =-274.80\times10^{3} A/m^{2}

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B=\frac{6.68\times10^{-5}}{8.14\times10^{-5}} =0.820 T

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B=\frac{u_{0}(I_w-I_c)}{2\pi d_3 } =\frac{2.011\times10^-5}{3.441\times10^{-4}} =0.0584 T

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