Question

Помогите пожалуйста с задачей!!! Заряд на каждом из двух последовательно соединенных конденсаторов емкостью С1=6 пкФ и С2=10 пкФ равен q=0,09 нКл. Определить напряжение: а) на батарее конденсаторов; б) на каждом конденсаторе.

Answer 1

Дано:
q =q1=q2 =0.09нКл
С1=18пФ
С2=10пФ
___________________________________
Найти: 
U - ?
U1 - ?
U2 - ?
_____________________________________

U = q/С = q (C1+C2) / C1C2 = 0.09*10^-9Кл * 28*10^-12Ф / 18*10*10^-12 * 10^-12 Ф*Ф =14В
С=С1С2 / (C1 + C2)
U1= q1/С1  = 0,09*10^-9 Кл / 18*10^-12Ф = 5В
U2=U - U1 = 9В

Answer 2

Problem Analysis

We have two capacitors, C1 and C2, connected in series. The total charge on both capacitors is given as q = 0.09 nC (nanocoulombs). We need to determine the voltage across the battery of the capacitors (V_battery) and the voltage across each capacitor (V_C1 and V_C2).

Solution

To solve this problem, we can use the formula relating charge (q), capacitance (C), and voltage (V) for a capacitor: q = C * V.

# Part (a): Voltage across the battery of the capacitors (V_battery)

The total charge on the capacitors is given as q = 0.09 nC. Since the capacitors are connected in series, the charge on each capacitor is the same. Therefore, we can write the equation as follows:

q = q1 = q2

Using the formula q = C * V, we can express the charge in terms of voltage for each capacitor:

q1 = C1 * V_C1 q2 = C2 * V_C2

Since q1 = q2, we can equate the two equations:

C1 * V_C1 = C2 * V_C2

We know the values of C1 and C2, so we can substitute them into the equation:

6 pF * V_C1 = 10 pF * V_C2

To find the voltage across the battery (V_battery), we need to sum the voltages across each capacitor:

V_battery = V_C1 + V_C2

# Part (b): Voltage across each capacitor (V_C1 and V_C2)

We can use the equation derived above to find the voltage across each capacitor.

Now, let's solve the equations to find the values of V_battery, V_C1, and V_C2.

Calculation

Given: C1 = 6 pF C2 = 10 pF q = 0.09 nC

Using the equation C1 * V_C1 = C2 * V_C2, we can solve for V_C1:

6 pF * V_C1 = 10 pF * V_C2

Dividing both sides by 6 pF:

V_C1 = (10 pF / 6 pF) * V_C2

Substituting the value of V_C1 in terms of V_C2 into the equation V_battery = V_C1 + V_C2:

V_battery = [(10 pF / 6 pF) * V_C2] + V_C2

Simplifying the equation:

V_battery = (10/6 + 1) * V_C2

Now, we can substitute the given values of C1, C2, and q into the equation q = C * V to find the value of V_C2:

0.09 nC = 10 pF * V_C2

Solving for V_C2:

V_C2 = (0.09 nC) / (10 pF)

Calculating the value of V_C2:

V_C2 = 0.009 V

Substituting the value of V_C2 into the equation for V_battery:

V_battery = (10/6 + 1) * 0.009 V

Calculating the value of V_battery:

V_battery = 0.015 V

Answer

The solutions to the problem are as follows: a) The voltage across the battery of the capacitors (V_battery) is 0.015 V. b) The voltage across each capacitor (V_C1 and V_C2) is 0.009 V.

Please let me know if you need any further assistance!