Question

1. a small drop of water, of mass ( 4.3\times10^{-9}\text{ kg} ), is suspended motionless by a uniform electric field of ( 9.2\times10^{2}\text{ n/c up} ). (a) is the charge on the drop positive or negative? explain. (b) find the number of extra electrons or protons on the drop.

Explanation:

Step1: Determine the charge sign

The electric force \(F_E = qE\) must balance the gravitational force \(F_g=mg\). Since the electric field \(E\) is upward and the gravitational force is downward (\(F_g = mg\) acts down), for \(F_E\) to be upward (to balance \(F_g\)), if \(E\) is upward (\(E>0\)), then \(q>0\) (because \(F_E = qE\)). So the charge is positive.

Step2: Calculate the charge \(q\)

Since \(F_E=F_g\), we have \(qE = mg\). Then \(q=\frac{mg}{E}\).
Substitute \(m = 4.3\times10^{-9}\space kg\), \(g = 9.8\space m/s^{2}\), and \(E=9.2\times 10^{2}\space N/C\)

$$ q=\frac{4.3\times 10^{-9}\times9.8}{9.2\times 10^{2}}=\frac{42.14\times 10^{-9}}{9.2\times 10^{2}}\approx4.58\times 10^{-11}\space C $$

Step3: Calculate the number of protons \(n\)

The charge of a proton \(e = 1.6\times 10^{-19}\space C\). Using \(q = ne\), then \(n=\frac{q}{e}\)

$$ n=\frac{4.58\times 10^{-11}}{1.6\times 10^{-19}}=\frac{4.58}{1.6}\times10^{8}\approx2.86\times 10^{8} $$

Answer:

(a) The charge is positive. Because the electric force \(F_E = qE\) must balance the gravitational force \(F_g=mg\) (down - ward). With \(E\) upward, \(q>0\) for \(F_E\) upward.

(b) The number of extra protons is approximately \(2.86\times 10^{8}\)