Suppose a van de Graaff generator builds a negative static charge, and a grounded conductor is placed near enough to it so that a 7.0 x 10-6 coulombs of negative charge arcs to the conductor. Calculate the number of electrons that are transferred.

Given Information:

Charge = q = 7.0x10⁻⁶ C

Required Information:

Number of electrons = n = ?

Number of electrons = 4.375x10¹³

Explanation:

We know that charge is given by

q = n*e

Where n is the number of electrons and e is the electronic charge that is 1.60x10⁻¹⁹ C

n = q/e

n = 7.0x10⁻⁶/1.60x10⁻¹⁹

n = 4.375x10¹³ electrons

Therefore, 4.375x10¹³ electrons were transferred.

Number of electrons transferred = 4.375 x 10^(13)

Explanation:

From coulombs law,

Mathematically, we can say that a charge is the number of electrons multiplied by the charge on 1 electron. Symbolically, it is given as;

Q = ne

where;

q is the symbol used to represent charge

n is a number of electrons;

e is a charge on 1 electron (1.6 × 10^(-19)C).

From the question,

Q = 7.0 x 10^(-6) coulombs of negative charge

Thus, making n the subject of the formula ;

n = Q/e = 7.0 x 10^(-6)/(1.6 × 10^(-19)) = 4.375 x 10^(13)

the answer is b, have negligible mass

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