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2 years ago

A 3.00 × 10^−9-coulomb test charge is placed near

Physics

2 answers:

viktelen [127]2 years ago

8 0

Answer:

Explanation:

Given:-

- The charge of the test particle q = 3.0 * 10^-9 C

- The force exerted by the metal sphere F = 6.0 * 10^-5 N

Find:-

The magnitude and direction of the electric field strength at this location?

Solution:-

- The relationship between the electrostatic force F exerted by the metal sphere on the test-charge and the Electric Field strength E at the position of test charge is given by:

F = E*q

- Using the data given we can determine E:

E = F / q

E = (6.0 * 10^-5) / (3.0 * 10^-9)

E = 20,000 N/C

- The direction of electric field is given by the net charge of the source ( metal sphere). The metal sphere is negative charge hence the direction of Electric Field strength E is directed towards the metal sphere.

patriot [66]2 years ago

3 0

Answer:

(2) 2.0×10⁴ N/C directed towards the sphere

Explanation:

Electric Field: This can be defined as the force per unit charge. The S.I unit of Electric Field is N/C.

The expression for electric Field is given as,

E = F/q...................... Equation 1

Where E = Electric Field, F = Force, q = charge.

Given: F = 6.0×10⁻⁵ N, q = 3×10⁻⁹ C

Substitute into equation equation 1

E = 6.0×10⁻⁵/(3×10⁻⁹)

E = 2.0×10⁴ N/C directed towards the sphere

Hence the right option is (2) 2.0×10⁴ N/C directed towards the sphere

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Answer:

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Now,buyantant force

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