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juin [17]
3 years ago
7

Problem 21-40a:

Physics
1 answer:
Greeley [361]3 years ago
8 0

Answer:

Part a)

\frac{F_e}{F_g} = 2.74 \times 10^{12}

Part b)

q = 3.37 \times 10^{-4} C

Explanation:

As we know that electric force on electric charge is given as

F = qE

here we have

q = 1.6 \times 10^{-19}C

E = 153 N/C

now force is given as

F = (1.6 \times 10^{-19})(153) = 2.45 \times 10^{-17} N

Gravitational force on electric charge near surface of earth is given as

F_g = mg

F_g = (9.1 \times 10^{-31})(9.81) = 8.93 \times 10^[-30} N

now the ratio of two forces is given as

\frac{F_e}{F_g} = \frac{2.45 \times 10^{-17}}{8.93 \times 10^{-30}}

\frac{F_e}{F_g} = 2.74 \times 10^{12}

Part b)

Now the ball is balanced by the electric force and the force of gravity on it

so here we have

F_g = qE

mg = qE

(5.25 \times 10^{-3})(9.81) = q(153)

here we have

q = 3.37 \times 10^{-4} C

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