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Andru [333]
3 years ago
8

A 6.0-v battery maintains the electrical potential difference between two parallel metal plates separated by 1.0 mm. what is the

electric field between the plates?
Physics
1 answer:
Gnoma [55]3 years ago
5 0
The voltage<span> difference between the two plates can be expressed in terms of the </span>work<span> done on a positive test charge q when it moves from the positive to the negative plate.</span><span>
E=V/d
where V is the voltage and d is the distance between the plates.
 
So,

E=6.0V/1mm= 6000 V/m. The electric field between the plates is 6000 V/m.</span>
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5 0
3 years ago
A 40.0 turn coil of wire of radius 3.0 cm is placed between the poles of an electromagnet. The field increases from 0 to 0.75 T
nadezda [96]

Answer:

Induced emf, V=3.76\times 10^{-4}\ V

Explanation:

We have,

Number of turns in the coil, N = 40

Radius of coil, r = 3 cm = 0.03 m

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It is required to find the magnitude of the induced emf in the coil if the field is perpendicular to the plane of the coil. The induced emf is given by :

\epsilon=\dfrac{d\phi}{dt}

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\epsilon=NA\dfrac{dB}{dt}\\\\\epsilon=\dfrac{40\times \pi (0.03)^2\times (0.75-0)}{225}\\\\\epsilon=3.76\times 10^{-4}\ V

So, the magnitude of induced emf is 3.76\times 10^{-4}\ V.

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