To solve this problem we will apply the concepts related to load balancing. We will begin by defining what charges are acting inside and which charges are placed outside.

PART A)

The charge of the conducting shell is distributed only on its external surface. The point charge induces a negative charge on the inner surface of the conducting shell:

$Q_{int}=-Q1=-1.9*10^{-6} C$ . This is the total charge on the inner surface of the conducting shell.

PART B)

The positive charge (of the same value) on the external surface of the conducting shell is:

$Q_{ext}=+Q_1=1.9*10^{-6} C$

The driver's net load is distributed through its outer surface. When inducing the new load, the total external load will be given by,

$Q_{ext, Total}=Q_2+Q_{ext}$

$Q_{ext, Total}=1.9+3.8$

$Q_{ext, Total}=5.7 \mu C$

5 0

3 years ago

what happens to the current in a circuit if the resitance of the components in the circuit is increased

Anit [1.1K]

Answer:

The current decreases.

Explanation:

Current and resistance are inversely proportional. The equation connecting current, resistance and voltage is $V = IR$ , where V is voltage, I is current and R is resistance.

Rearranging this equation, you get:

$I = \frac{V}{R}$

and

$R = \frac{V}{I}$

If the value of voltage in both equations remains constant, and the value of R decreases, the value of I will increase. Conversely, if in the second equation $R = \frac{V}{I}$ , the value of V remains constant the value of I decreases, then the value of R, resistance will increase.

Thus, it can be seen that the current will decrease as resistance increases and vice versa.

7 0

2 years ago

HELP ME PLEASE! according ohm's law if the resistance remains constant what would happen if you decrease the amount of voltage b

SpyIntel [72]

B.
technically it would depend if the resistors were in series or parallel but B is the answer.

7 0

3 years ago