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Q64PE

Expert-verifiedFound in: Page 667

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**(a) Calculate the electric field strength near a 10.0 cm diameter conducting sphere that has 1.00 C of excess charge on it. (b) What is unreasonable about this result? (c) Which assumptions are responsible?**

(a) The strength of the electric field is $3.6\times {10}^{12}N/C$ . (b) The field is too large to be in air. (c) 1.00 charge is excessive.

- Diameter of a sphere is

- 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,

$E=\frac{F}{{q}_{\circ}}$

Here, *F* is the electrostatic force and ${q}_{\circ}$ is the test charge.

The formula for electric field strength is,

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

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

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

role="math" localid="1653734461721" $E=\frac{\left(9\times {10}^{9}N-{m}^{2}/{C}^{2}\right)\times \left(1.00C\right)}{{\left(\frac{10.0cm}{2}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{\left(9\times {10}^{9}N-{m}^{2}/{C}^{2}\right)\times \left(1.00C\right)}{{\left[\left(\frac{10.0cm}{2}\right)\times \left(\frac{{10}^{-2}m}{1cm}\right)\right]}^{2}}\phantom{\rule{0ex}{0ex}}=3.6\times {10}^{12}N/C$

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

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

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

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

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