Answer:
The electric field is [tex]E = 7.5 *10^{6} \ N/C[/tex]
Explanation:
From the question we are told that
The radius of the metal sphere is [tex]R = 2.0 \ cm = 0.02 \ m[/tex]
The excess charge which the metal sphere carries is [tex]q = 3.0 \mu C = 3.0*10^{-6} \ C[/tex]
The distance of the position being to the center is [tex]D = 6.0 \ cm = 0.06 \ m[/tex]
The coulomb constant is [tex]k =9*10^{9} \ N \cdot m^2 /C^2[/tex]
Generally the electric field is mathematically represented as
[tex]E = \frac{k * q}{D^2}[/tex]
substituting values
[tex]E = \frac{9*10^{9} * 30.*10^{-6}}{(0.06)^2}[/tex]
[tex]E = 7.5 *10^{6} \ N/C[/tex]
The electric field from the center of the sphere will be "7.5 × 10⁶ N/C".
According to the question,
Metal sphere's radius, R = 2.0 cm or,
= 0.02 m
The excess charge, q = 3.0 μC or,
= 3.0 × 10⁻⁶ C
Position distance, D = 6.0 cm or,
= 0.06 cm
Coulomb constant, k = 9 × 10⁹ N.m²/C²
We know the relation,
Electric field, E = [tex]\frac{k\times q}{D^2}[/tex]
By substituting the values, we get
= [tex]\frac{9\times 10^9\times 3.0\times 10^{-6}}{(0.06)^2}[/tex]
= [tex]\frac{9\times 10^9\times 3.0\times 10^{-6}}{0.0036}[/tex]
= 7.5 × 10⁶ N/C
Thus the approach above is correct.
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