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E = q/(4PIEoR^2) plugged and chugged and got 3.34E-9 which isn't right of course, where did i screw up?

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In summary, the conversation discusses an unknown charge on a conducting solid sphere with a radius of 10 cm. The E-field at a distance of 15 cm from the center of the sphere is 3.0E3 N/c and directed radially inward. The net charge on the sphere is calculated using the formula E = q/(4PIEoR^2), with a resulting value of 3.34E-9. However, there is a discrepancy in the calculated answer and it is suggested that the arithmetic may have been done incorrectly. The speaker suggests using the formula E= (kQ)/r^2 to calculate the charge, taking into account that the charge will be negative since the field is directed inward.

- #1

- 1,629

- 1

E = q/(4PIEoR^2) plugged and chugged and got 3.34E-9 which isn't right of course, where did i screw up?

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[itex]

E= \frac{kQ}{r^2}

[/itex]

You just need to fill in the values to calculate Q.Since the field is inwards, the charge will be negative.

BJ

The formula for finding the net charge of a sphere is Q = 4πε_{0}R^{2}Φ, where Q is the net charge in Coulombs, ε_{0} is the permittivity of free space, R is the radius of the sphere in meters, and Φ is the electric flux in volts.

Yes, the net charge of a sphere can be negative. This means that the sphere has an overall excess of electrons, giving it a negative charge.

The net charge of a sphere is affected by the amount of charge present on the surface of the sphere, the radius of the sphere, and the electric flux passing through the sphere.

The net charge of a sphere affects the strength and direction of its electric field. A positive net charge will create an outward electric field, while a negative net charge will create an inward electric field.

Yes, the net charge of a sphere can be zero if there is an equal amount of positive and negative charge present on the surface of the sphere, or if the electric flux passing through the sphere is zero.

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