91Ó°ÊÓ

• Charge on sphere is 1.00 C

The electric field is defined as the force experienced by a unit positive test charge due to the presence of another charge.The expression for the electric field is,

$E=\frac{F}{q_{∘}}$

Here, F is the electrostatic force and $q_{∘}$is the test charge.

The formula for electric field strength is,

$E=\frac{Kq}{{\left({\displaystyle \frac{d}{2}}\right)}^2}$

Here, K is the electrostatic force constant, q is the charge on the conducting sphere, and d is the diameter of the sphere.

Substitute $K=9×{10}^{9}N-m^2/C^2,q=1.00C,d=10.0cm,$

role="math" localid="1653734461721" $$\begin{gathered} E=\frac{\left(9×{10}^{9}N-m^2/C^2\right)×\left(1.00C\right)}{{\left({\displaystyle \frac{10.0cm}{2}}\right)}^2} \\ =\frac{\left(9×{10}^{9}N-m^2/C^2\right)×\left(1.00C\right)}{{\left[\left({\displaystyle \frac{10.0cm}{2}}\right)×\left({\displaystyle \frac{{10}^{-2}m}{1cm}}\right)\right]}^2} \\ =3.6×{10}^{12}N/C \end{gathered}$$

Hence, the value of the electric field strength is $3.6×{10}^{12}N/C$.

The dielectric breakdown occurs at $3×{10}^{6}N/C$, which is much less than the field outside this sphere. Sparks would occur before reaching the field strength of $3.6×{10}^{12}N/C$.

Hence, the field is too large to be in air.

The responsible assumption is that, the 1.00 C charge is excessive.