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lianna [129]
3 years ago
7

Calculate the electric field strength at a point at which a test charge of 0.30 coulombs experiences a force of 5.0 newtons.

Physics
1 answer:
SVEN [57.7K]3 years ago
5 0

The strength of electric field E is 17 N / C.

<u />

<u>Explanation:</u>

Electric field strength is defined as the force per unit charge acting at a point in the given field. The equation for the strength of the electric field is given by

                     E = F / q

where E represents the electric field strength,

           F represents the force in newton,

           q represents the charge in coulomb.

Given the charge q = 0.30 coulombs

                   force F = 5.0 N

Electric field strength E = force / charge

                                        = 5.0 / 0.30

                                    E  = 16.66 = 17 N / C.

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The magnitude of the magnetic field inside the solenoid is 6.46 \times 10^{-3} \ T.

The given parameters;

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The magnitude of the magnetic field inside the solenoid is calculated as;

B = \mu_0 nI\\\\B = \mu_o(\frac{ N}{L} )I\\\\

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B = (4\pi \times 10^{-7}) \times (\frac{1300}{0.91} ) \times 3.6\\\\B = 6.46 \times 10^{-3} \ T

Thus, the magnitude of the magnetic field inside the solenoid is 6.46 \times 10^{-3} \ T.

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