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2 years ago

Compared to the weak nuclear force the electromagnetic force

Physics

2 answers:

Crazy boy [7]2 years ago

8 0

Explanation:

Weak nuclear force:

The interaction between the subatomic particles is called weak nuclear force.

The weak nuclear force is one of the four fundamental forces.

The weak nuclear force is effective at very short distance.

The range and relative strength of weak nuclear force is 10⁻¹⁸ m and 10²⁵ with respect to gravitational force respectively

Deuterium is formed due to the fusion of protons and neutrons under the action the weak force.

Example : Beta decay

Electromagnetic force:

The interaction between the charged particles is called electromagnetic force.

The electromagnetic force is one of the four fundamental forces.

The electromagnetic force is effective at long range distance.

The range and relative strength of electromagnetic force is infinity and 10³⁶ with respect to gravitational force respectively

Example : light

kkurt [141]2 years ago

8 0

Answer:

The answer is A. on edgen.

Explanation:

Compared to the weak nuclear force, the electromagnetic force A. is Stronger

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2 years ago

There are two parallel conductive plates separated by a distance d and zero potential. Calculate the potential and electric fiel

taurus [48]

Answer:

The total electric potential at mid way due to 'q' is $\frac{q}{4\pi\epsilon_{o}d}$

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The electric potential at a distance d due to 'Q' is:

$V = \frac{1}{4\pi\epsilon_{o}}.\frac{Q}{d}$

Now, for the Electric potential for the two plates A and B at midway between the plates due to 'q':

For plate A,

$V_{A} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{\frac{d}{2}}$

Similar is the case with plate B:

$V_{B} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{\frac{d}{2}}$

Since the electric potential is a scalar quantity, the net or total potential is given as the sum of the potential for the two plates:

$V_{total} = V_{A} + V_{B} = \frac{1}{4\pi\epsilon_{o}}.q(\frac{1}{\frac{d}{2}} + \frac{1}{\frac{d}{2}}$

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The Electric field due to charge Q at a distance is given by:

$\vec{E} = \frac{1}{4\pi\epsilon_{o}}.\frac{Q}{d^{2}}$

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$\vec{E_{A}} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{(\frac{d}{2})^{2}}$

Similarly, for plate B:

$\vec{E_{B}} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{(\frac{d}{2})^{2}}$

Both the fields for plate A and B are due to charge 'q' and as such will be equal in magnitude with direction of fields opposite to each other and hence cancels out making net Electric field zero.

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3 years ago

Read 2 more answers