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3 years ago

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State and explain the principle of superposition of electrostatic force

1 answer:

ella [17]3 years ago

8 0

Answer:

The principle of superposition states that when a number of charges are interacting, the net electrostatic force a given charge is the vector sum of the forces exerted on it due to all other charges. The force between two charges is not affected by the presence of other charges.

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If a proton and an electron are released when they are 4.00 x 10^-10 m apart, find the initial acceleration of each of them.

SVEN [57.7K]

Answer:

For proton

$a=8.8\times 10^{17}\ m/s^2$

For electron

$a=1.5\times 10^{21}\ m/s^2$

Explanation:

We know that

The mass of electron

$m=9.1\times 10^{-31}\ kg$

The mass of proton

$m=1.67\times 10^{-27}\ kg$

Charge on electron and proton

q₁=q₂=q

$q=1.6\times 10^{-19}\ C$

Electrostatics force

$F=K\dfrac{q_1q_2}{r^2}$

Now by putting the values

$F=9\times 10^9\times \dfrac{1.6\times 10^{-19}\times 1.6\times 10^{-19}}{(4\times 10^{-10})^2}$

$F=1.44\times 10^{-9}\ N$

For proton

F = m a

a =F/m

$a=\dfrac{1.4\times 10^{-9}}{1.67\times 10^{-27}}\ m/s^2$

$a=8.8\times 10^{17}\ m/s^2$

For electron

$a=\dfrac{1.4\times 10^{-9}}{9.1\times 10^{-31}}\ m/s^2$

$a=1.5\times 10^{21}\ m/s^2$

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A charged particle is moving in a uniform magnetic field at a speed of 8.2Ã10^3 m/s in a direction 87Â° from the direction of th

77julia77 [94]

Answer:

Magnetic field, B = 0.042 T

Explanation:

It is given that,

Speed of charged particle, $v=8.2\times 10^3\ m/s$

Angle between velocity and the magnetic field, $\theta=87$

Charge, $q=5.7\ \mu C=5.7\times 10^{-6}\ C$

Magnetic force, F = 0.002 N

The magnetic force is given by :

$F=qvB\ sin\theta$

B is the magnetic field

$B=\dfrac{F}{qv\ sin\theta}$

$B=\dfrac{0.002}{5.7\times 10^{-6}\times 8.2\times 10^3\times sin(87)}$

B = 0.042 T

So, the strength of the magnetic field is 0.042 Tesla. Hence, this is the required solution.

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3 years ago