(a) 6.95 x 10⁻⁸ C
(b) 6.25N/C
The electric field (E) on a point charge, Q, is given by;
E = k x Q / r² ---------------(i)
Where;
k = constant = 8.99 x 10⁹ N m²/C²
r = distance of the charge from a reference point.
Given from the question;
E = 10000N/C
r = 0.250m
Substitute these values into equation(i) as follows;
10000 = 8.99 x 10⁹ x Q / (0.25)²
10000 = 8.99 x 10⁹ x Q / (0.0625)
10000 = 143.84 x 10⁹ x Q
Solve for Q;
Q = 10000/(143.84 x 10⁹)
Q = 0.00695 x 10⁻⁵C
Q = 6.95 x 10⁻⁸ C
The magnitude of the charge is 6.95 x 10⁻⁸ C
(b) To get how large the field (E) will be at r = 10.0m, substitute these values including Q = 6.95 x 10⁻⁸ C into equation (i) as follows;
E = k x Q / r²
E = 8.99 x 10⁹ x 6.95 x 10⁻⁸ / 10²
E = 8.99 x 10⁹ x 6.95 x 10⁻⁸ / 100
E = 6.25N/C
Therefore, at 10.0m, the electric field will be just 6.25N/C
The magnitude of the electric field is proportional to the charge of the particle. The magnitude of the given charge is [tex]6.95 \times 10^{-8}\rm \ C[/tex].
The magnitude of the electric field is proportional to the charge of the particle.
[tex]E = k \times \dfrac {Q }{r^2}[/tex]
Where,
E - electric field = 10000 N/C
k - onstant = 8.99 x 10⁹ N m²/C²
Q - charge
r - distance = 0.25 m
Put the values formula,
[tex]10000 = 8.99 \times 10^9\times \dfrac { Q}{ (0.25)^2}\\\\Q = 6.95 \times 10^{-8}\rm \ C[/tex]
Therefore, the magnitude of the given charge is [tex]6.95 \times 10^{-8}\rm \ C[/tex].
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