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emmainna [20.7K]
3 years ago
13

Which changes will decrease the electric force between two positively charged objects? Check all that apply.

Physics
2 answers:
zaharov [31]3 years ago
8 0

Answer:

Moving them farther apart

Explanation:

The electric field between the two charges Q and q separated by a distance r is given by

E = \frac{K\times Q\times q}{r^{2}}

It shows that the electric field is inversely proportional to the square of the distance between two charges.

So, as the distance between two charges increases, the electric filed between the two charges decreases.

GaryK [48]3 years ago
8 0

Answer:

1. moving them farther apart

2. Adding electrons

Explanation:

The electric force between two positively charged particle is given by :

F=\dfrac{kq_1q_2}{r^2}

Where

k is the electrostatic constant

r is the distance between charges

The electric force is inversely proportional to the distance between them. So, the electric force between two positively charged objects will decrease when the charged particles move father apart. Also, on adding electrons, the magnitude of charges will decrease.

Hence, the correct options are (b) and (c).

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A capacitor with initial charge q0 is discharged through a resistor. a) In terms of the time constant τ, how long is required fo
-BARSIC- [3]

Answer:

It would take \tau(\ln 9 - \ln 8) time for the capacitor to discharge from q_0 to \displaystyle \frac{8}{9} \, q_0.

It would take \tau(\ln 9 - \ln 7) time for the capacitor to discharge from q_0 to \displaystyle \frac{7}{9}\, q_0.

Note that \ln 9 = 2\,\ln 3, and that\ln 8 = 3\, \ln 2.

Explanation:

In an RC circuit, a capacitor is connected directly to a resistor. Let the time constant of this circuit is \tau, and the initial charge of the capacitor be q_0. Then at time t, the charge stored in the capacitor would be:

\displaystyle q(t) = q_0 \, e^{-t / \tau}.

<h3>a)</h3>

\displaystyle q(t) = \left(1 - \frac{1}{9}\right) \, q_0 = \frac{8}{9}\, q_0.

Apply the equation \displaystyle q(t) = q_0 \, e^{-t / \tau}:

\displaystyle \frac{8}{9}\, q_0 = q_0 \, e^{-t/\tau}.

The goal is to solve for t in terms of \tau. Rearrange the equation:

\displaystyle e^{-t/\tau} = \frac{8}{9}.

Take the natural logarithm of both sides:

\displaystyle \ln\, e^{-t/\tau} = \ln \frac{8}{9}.

\displaystyle -\frac{t}{\tau} = \ln 8 - \ln 9.

t = - \tau \, \left(\ln 8 - \ln 9\right) = \tau(\ln 9 - \ln 8).

<h3>b)</h3>

\displaystyle q(t) = \left(1 - \frac{1}{9}\right) \, q_0 = \frac{7}{9}\, q_0.

Apply the equation \displaystyle q(t) = q_0 \, e^{-t / \tau}:

\displaystyle \frac{7}{9}\, q_0 = q_0 \, e^{-t/\tau}.

The goal is to solve for t in terms of \tau. Rearrange the equation:

\displaystyle e^{-t/\tau} = \frac{7}{9}.

Take the natural logarithm of both sides:

\displaystyle \ln\, e^{-t/\tau} = \ln \frac{7}{9}.

\displaystyle -\frac{t}{\tau} = \ln 7 - \ln 9.

t = - \tau \, \left(\ln 7 - \ln 9\right) = \tau(\ln 9 - \ln 7).

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