Answer:
Charge, q = 1.15 C
Explanation:
It is given that,
Mass of sphere, m = 441 g = 0.441 kg
Electric field, E = 5 N/C
Due to this field, the sphere suddenly acquires a horizontal acceleration of 13.0 m/s² such that,
[tex]ma=qE[/tex]
[tex]q=\dfrac{ma}{E}[/tex]
[tex]q=\dfrac{0.441\ kg\times 13\ m/s^2}{5\ N/C}[/tex]
q = 1.146 C
or
q = 1.15 C
So, the charge carried by the small sphere is 1.15 C. Hence, this is the required solution.
The electric field accounts for the unbalanced motion of the sphere. The electrostatic forces balance the linear force, for which the charge acquired by the sphere is 1.146 C.
The region or space where a charged particle experiences the electrostatic force of attraction or repulsion, is known as an electric field.
Given data:
The mass of the small sphere is, m = 441 g = 0.441 kg.
The magnitude of the electric field is, E = 5.00 N/C.
The magnitude of horizontal acceleration is, a = 13.0 m/s².
Clearly, under the influence of an electric field, the electrostatic force balances the linear force. Then,
Fe = FL
[tex]qE = ma[/tex]
Here,
q is the magnitude of charge.
Solving as,
[tex]q \times 5.0 = 0.441 \times 13\\\\q = \dfrac{0.441 \times 13}{5.0}\\\\q = 1.146 \;\rm C[/tex]
Thus, we can conclude that the magnitude of charge carried by the sphere is 1.146 C.
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