ByMpRoduct
What electric charge does a sphere with a surface area of 7500 cm² carry if
surface has a field strength of 50000 V/m? How many electrons must
are therefore located on the surface.
My solution is Q=3.32*10^-7 C and the number of electrons = 2.07*10^12 electrons. Thank you in advance.
(1 votes)
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Right.
Q = E * A
Q represents the charge, E represents the field strength and A represents the surface of the ball.
There is a field strength of 50000 V/m and a surface of 7500 cm2. To convert the surface into square meters, we divide it by 10000:
A = 7500 cm2 / 10000 = 0,75 m2
Now we put the values into the formula:
Q = 50000 V/m * 0,75 m2 = 37500 C
The ball thus carries a charge of 37500 Coulomb.
To find out how many electrons are on the surface, we use the elementary charge e, which is about 1.6*10^-19 Coulomb. We share the total charge through the elementary charge:
Number of electrons = Q / e = 37500 C / (1.6 * 10^-19 C) ≅ 2.34 * 10^22 electrons
Accordingly, there are approximately 2.34*10^22 electrons on the surface of the sphere.
Is unfortunately already a little bit here, so I do not give a 100% guarantee for the answer 🙂
Q = E * A
There is no factor Epsilon 0
Thank you for your answer. Your solution is wrong. Did you do this with ChatGPT?