When a honeybee flies through the air it develops a charge of +20pC. How many electrons did it lose in the process of acquiring this charge?

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When a honeybee flies through the air it develops a charge of +20pC. How many electrons did it lose in the process of acquiring this charge?

Express answer as a number.

Total charge acquired= +20 pC = 20*10^-12 C

Since the honeybee was neutral before, the total charge acquired would be equal to n*charge of electron, where n is number of electrons lost

charge on each electron = 1.6*10^-19 C

n*1.6*10^-19 = 20*10-12

n=12.5*107

Number of electrons lost = 12.5*10^7

When a honeybee flies through the air it develops a charge of +19pC. How many electrons did it lose in the process of acquiring this charge?

Express answer as a number.

charge , Q = + 17 pC = 17 * 10^-12 C

magnitude of charge on electron ,q = 1.6 * 10^-19 C

the number of electron , n = Q/q

n = 17 * 10^-12 /( 1.6 * 10^-19) electrons

n = 1.06 * 10^8 electrons

When a honeybee flies through the air it develops a charge of +18pC. How many electrons did it lose in the process of acquiring this charge?

Express answer as a number.

As we know,

Charge on an electron = -1.6×10^(-19) C

According to question when a honeybee flies in the sky it developes a charge = +18pC

Which is equal to +18×10^(-12) C

We can say, when one electron is lost by any body then +1.6×10^(-19) C charge is developed on that body.

So to develope a charge of +18×10^(-12) C number of electrons to be lost by the body is = 18×10^(-12) ÷ 1.6×10^(-19) = 112500000 electrons

Therefore,

honeybee losses 112500000 (11.25×10^7) electrons to get charge of 18×10^(-12) C.

When a honeybee flies through the air it develops a charge of +17pC. How many electrons did it lose in the process of acquiring this charge?

Express answer as a number.

charge , Q = + 17 pC = 17 * 10^-12 C

magnitude of charge on electron ,q = 1.6 * 10^-19 C

the number of electron , n = Q/q

n = 17 * 10^-12 /( 1.6 * 10^-19) electrons

n = 1.06 * 10^8 electrons

When a honeybee flies through the air it develops a charge of +16pC. How many electrons did it lose in the process of acquiring this charge?

Express answer as a number.

Q=+16 pC→16×10^-12 C

n = Q/e

n = 1.0 * 10^8 electrons = 10^8.

When a honeybee flies through the air it develops a charge of +15pC. How many electrons did it lose in the process of acquiring this charge?

Express answer as a number.

n = 93750000 electrons

When a honeybee flies through the air, it develops a chargeof +15pC. How many electrons did it lose in the processof acquiring this charge?

When a honeybee flies through the air it develops a charge of +14pC. How many electrons did it lose in the process of acquiring this charge?

Express answer as a number.

n = 8×75^7 electrons

When a honeybee flies through the air, it develops a charge of+16pC. How many electrons did it lose in the process of acquiringthis charge? Express your answer as a number of electrons.

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