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A solid sphere of radius R = 5 cm is made of non-conducting material and carries a total negative charge Q = -12 C. The charge is uniformly distributed throughout the interior of the sphere.

What is the magnitude of the electric potential V at a distance r = 30 cm from the center of the sphere, given that the potential is zero at r = [infinity] ?

Sagot :

Cjrodriguez89idn

Answer:

V= -3.6*10⁻¹¹ V

Explanation:

  • Since the charge is uniformly distributed, outside the sphere, the electric field is radial (due to symmetry), so applying Gauss' Law to a spherical surface at r= 30 cm, we can write the following expression:

      [tex]E* A = \frac{Q}{\epsilon_{0} } (1)[/tex]

  • At r= 0.3 m the spherical surface can be written as follows:

       [tex]A = 4*\pi *r^{2} = 4*\pi *(0.3m)^{2} (2)[/tex]

  • Replacing (2) in (1) and solving for E, we have:

      [tex]E = \frac{Q}{4*\pi *\epsilon_{0}*r^{2} } = \frac{(9e9N*m2/C2)*(-12C)}{(0.3m)^{2} y} (3)[/tex]

  • Since V is the work done on the charge by the field, per unit charge, in this case, V is simply:
  • V = E. r (4)
  • Replacing (3) in (4), we get:

       [tex]V =E*r = E*(0.3m) = \frac{(9e9N*m2/C2)*(-12C)}{(0.3m)} = -3.6e11 V (5)[/tex]

  • V = -3.6*10¹¹ Volts.
Annykslidn

The electrical potential module will be [tex]-3.6*10^-^1^1 V[/tex]

We can arrive at this answer as follows:

  • To answer this, we owe Gauss's law. This is because the charge is evenly distributed across the sphere. This will be done as follows:

[tex]E*A=\frac{Q}{^E0} \\\\\\A=4*\pi*r^2[/tex]

  • Solving these equations will have:

[tex]E=\frac{Q}{4*\pi*^E0*r^2} \\E= \frac{(9e9N*m2/c2)*(-12C)}{(0.3m)^2y}[/tex]

  • As we can see, the electric potential is carried out on the field charge. In this case, using the previous equations, we can calculate the value of V as follows:

[tex]V=E*r\\V=E*0.3m= \frac{(9e9N*m^2/C2)*(-12C)}{0.3m} \\V= -3.6*10^-^1^1 V.[/tex]

More information about Gauss' law at the link:

https://brainly.com/question/14705081

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