(a) The time required to lose 90% of its initial charge and energy

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Answer 1

Question

Chapter 18, Problem 75QAP

To determine

(a)

The time required to lose 90% of its initial charge and energy

Expert Solution & Answer

Explanation of Solution

Given:

Capacitor, C=12.5μF=12.5×10−6F

Resistor, R=75.0Ω

Voltage of the battery, V=50.0V

Formula used:

Charge on the capacitor in time t is given by,
Q=C×V

Where,
Q= charge on the capacitor
C= capacitance
V = voltage
Energy stored in the capacitor is,
E=12CV2

Calculation:

Charge on the capacitor is given by,
Q0=C×VQ0=12.5×10−6×50.0Q0=625×10−6C

The new charge is given by.

Q=Q0−0.9Q0Q=0.1Q0Q=0.1×625×10−6Q=62.5×10−6C

Thus, new voltage is,
Vn=QCVn=62.5× 10 −612.5× 10 −6Vn=5V

Voltage of the capacitor in time t is given by,
Vn=V(1−e −t RC)5=50.0×(1−e −t 75.0×12.5× 10 −6 )1−e −t 75.0×12.5× 10 −6 =0.1e −t 75.0×12.5× 10 −6 =0.9−0.04575=−t75.0×12.5× 10 −6t=9.8775×10−5s

Energy stored in the capacitor is,

E=12CV2E=0.5×12.5×10−6×50.0×50.0E=0.01562J

Thus, new charge is,
En=0.1×0.015625En=1.5625×10−3J

So, new voltage is,
En=12CVn2Vn= 2× E n CVn=15.81V

Voltage of the capacitor in time t is given by,
Vn=V(1−e −t RC)15.81=50.0×(1−e −t 75.0×12.5× 10 −6 )1−e −t 75.0×12.5× 10 −6 =0.3162e −t 75.0×12.5× 10 −6 =0.6838−0.3800=−t75.0×12.5× 10 −6t=3.56×10−4s

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Students have asked these similar questions

Example Two charges, one with +10 μC of charge, and another with - 7.0 μC of charge are placed in line with each other and held at a fixed distance of 0.45 m. Where can you put a 3rd charge of +5 μC, so that the net force on the 3rd charge is zero?

* Coulomb's Law Example Three charges are positioned as seen below. Charge 1 is +2.0 μC and charge 2 is +8.0μC, and charge 3 is - 6.0MC. What is the magnitude and the direction of the force on charge 2 due to charges 1 and 3? 93 kq92 F == 2 r13 = 0.090m 91 r12 = 0.12m 92 Coulomb's Constant: k = 8.99x10+9 Nm²/C² ✓

Make sure to draw a Free Body Diagram as well

Answer 2

Question

Chapter 18, Problem 75QAP

To determine

(a)

The time required to lose 90% of its initial charge and energy

Expert Solution & Answer

Explanation of Solution

Given:

Capacitor, C=12.5μF=12.5×10−6F

Resistor, R=75.0Ω

Voltage of the battery, V=50.0V

Formula used:

Charge on the capacitor in time t is given by,
Q=C×V

Where,
Q= charge on the capacitor
C= capacitance
V = voltage
Energy stored in the capacitor is,
E=12CV2

Calculation:

Charge on the capacitor is given by,
Q0=C×VQ0=12.5×10−6×50.0Q0=625×10−6C

The new charge is given by.

Q=Q0−0.9Q0Q=0.1Q0Q=0.1×625×10−6Q=62.5×10−6C

Thus, new voltage is,
Vn=QCVn=62.5× 10 −612.5× 10 −6Vn=5V

Voltage of the capacitor in time t is given by,
Vn=V(1−e −t RC)5=50.0×(1−e −t 75.0×12.5× 10 −6 )1−e −t 75.0×12.5× 10 −6 =0.1e −t 75.0×12.5× 10 −6 =0.9−0.04575=−t75.0×12.5× 10 −6t=9.8775×10−5s

Energy stored in the capacitor is,

E=12CV2E=0.5×12.5×10−6×50.0×50.0E=0.01562J

Thus, new charge is,
En=0.1×0.015625En=1.5625×10−3J

So, new voltage is,
En=12CVn2Vn= 2× E n CVn=15.81V

Voltage of the capacitor in time t is given by,
Vn=V(1−e −t RC)15.81=50.0×(1−e −t 75.0×12.5× 10 −6 )1−e −t 75.0×12.5× 10 −6 =0.3162e −t 75.0×12.5× 10 −6 =0.6838−0.3800=−t75.0×12.5× 10 −6t=3.56×10−4s

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Answer 3

Question

Chapter 18, Problem 75QAP

To determine

(a)

The time required to lose 90% of its initial charge and energy

Expert Solution & Answer

Explanation of Solution

Given:

Capacitor, C=12.5μF=12.5×10−6F

Resistor, R=75.0Ω

Voltage of the battery, V=50.0V

Formula used:

Charge on the capacitor in time t is given by,
Q=C×V

Where,
Q= charge on the capacitor
C= capacitance
V = voltage
Energy stored in the capacitor is,
E=12CV2

Calculation:

Charge on the capacitor is given by,
Q0=C×VQ0=12.5×10−6×50.0Q0=625×10−6C

The new charge is given by.

Q=Q0−0.9Q0Q=0.1Q0Q=0.1×625×10−6Q=62.5×10−6C

Thus, new voltage is,
Vn=QCVn=62.5× 10 −612.5× 10 −6Vn=5V

Voltage of the capacitor in time t is given by,
Vn=V(1−e −t RC)5=50.0×(1−e −t 75.0×12.5× 10 −6 )1−e −t 75.0×12.5× 10 −6 =0.1e −t 75.0×12.5× 10 −6 =0.9−0.04575=−t75.0×12.5× 10 −6t=9.8775×10−5s

Energy stored in the capacitor is,

E=12CV2E=0.5×12.5×10−6×50.0×50.0E=0.01562J

Thus, new charge is,
En=0.1×0.015625En=1.5625×10−3J

So, new voltage is,
En=12CVn2Vn= 2× E n CVn=15.81V

Voltage of the capacitor in time t is given by,
Vn=V(1−e −t RC)15.81=50.0×(1−e −t 75.0×12.5× 10 −6 )1−e −t 75.0×12.5× 10 −6 =0.3162e −t 75.0×12.5× 10 −6 =0.6838−0.3800=−t75.0×12.5× 10 −6t=3.56×10−4s

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Answer 4

Question

Chapter 18, Problem 75QAP

To determine

(a)

The time required to lose 90% of its initial charge and energy

Expert Solution & Answer

Explanation of Solution

Given:

Capacitor, C=12.5μF=12.5×10−6F

Resistor, R=75.0Ω

Voltage of the battery, V=50.0V

Formula used:

Charge on the capacitor in time t is given by,
Q=C×V

Where,
Q= charge on the capacitor
C= capacitance
V = voltage
Energy stored in the capacitor is,
E=12CV2

Calculation:

Charge on the capacitor is given by,
Q0=C×VQ0=12.5×10−6×50.0Q0=625×10−6C

The new charge is given by.

Q=Q0−0.9Q0Q=0.1Q0Q=0.1×625×10−6Q=62.5×10−6C

Thus, new voltage is,
Vn=QCVn=62.5× 10 −612.5× 10 −6Vn=5V

Voltage of the capacitor in time t is given by,
Vn=V(1−e −t RC)5=50.0×(1−e −t 75.0×12.5× 10 −6 )1−e −t 75.0×12.5× 10 −6 =0.1e −t 75.0×12.5× 10 −6 =0.9−0.04575=−t75.0×12.5× 10 −6t=9.8775×10−5s

Energy stored in the capacitor is,

E=12CV2E=0.5×12.5×10−6×50.0×50.0E=0.01562J

Thus, new charge is,
En=0.1×0.015625En=1.5625×10−3J

So, new voltage is,
En=12CVn2Vn= 2× E n CVn=15.81V

Voltage of the capacitor in time t is given by,
Vn=V(1−e −t RC)15.81=50.0×(1−e −t 75.0×12.5× 10 −6 )1−e −t 75.0×12.5× 10 −6 =0.3162e −t 75.0×12.5× 10 −6 =0.6838−0.3800=−t75.0×12.5× 10 −6t=3.56×10−4s

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Answer 5

Chapter 18, Problem 75QAP

To determine

(a)

The time required to lose 90% of its initial charge and energy

Expert Solution & Answer

Explanation of Solution

Given:

Capacitor, C=12.5μF=12.5×10−6F

Resistor, R=75.0Ω

Voltage of the battery, V=50.0V

Formula used:

Charge on the capacitor in time t is given by,
Q=C×V

Where,
Q= charge on the capacitor
C= capacitance
V = voltage
Energy stored in the capacitor is,
E=12CV2

Calculation:

Charge on the capacitor is given by,
Q0=C×VQ0=12.5×10−6×50.0Q0=625×10−6C

The new charge is given by.

Q=Q0−0.9Q0Q=0.1Q0Q=0.1×625×10−6Q=62.5×10−6C

Thus, new voltage is,
Vn=QCVn=62.5× 10 −612.5× 10 −6Vn=5V

Voltage of the capacitor in time t is given by,
Vn=V(1−e −t RC)5=50.0×(1−e −t 75.0×12.5× 10 −6 )1−e −t 75.0×12.5× 10 −6 =0.1e −t 75.0×12.5× 10 −6 =0.9−0.04575=−t75.0×12.5× 10 −6t=9.8775×10−5s

Energy stored in the capacitor is,

E=12CV2E=0.5×12.5×10−6×50.0×50.0E=0.01562J

Thus, new charge is,
En=0.1×0.015625En=1.5625×10−3J

So, new voltage is,
En=12CVn2Vn= 2× E n CVn=15.81V

Voltage of the capacitor in time t is given by,
Vn=V(1−e −t RC)15.81=50.0×(1−e −t 75.0×12.5× 10 −6 )1−e −t 75.0×12.5× 10 −6 =0.3162e −t 75.0×12.5× 10 −6 =0.6838−0.3800=−t75.0×12.5× 10 −6t=3.56×10−4s

Answer 6

Chapter 18, Problem 75QAP

To determine

(a)

The time required to lose 90% of its initial charge and energy

Expert Solution & Answer

Explanation of Solution

Given:

Capacitor, C=12.5μF=12.5×10−6F

Resistor, R=75.0Ω

Voltage of the battery, V=50.0V

Formula used:

Charge on the capacitor in time t is given by,
Q=C×V

Where,
Q= charge on the capacitor
C= capacitance
V = voltage
Energy stored in the capacitor is,
E=12CV2

Calculation:

Charge on the capacitor is given by,
Q0=C×VQ0=12.5×10−6×50.0Q0=625×10−6C

The new charge is given by.

Q=Q0−0.9Q0Q=0.1Q0Q=0.1×625×10−6Q=62.5×10−6C

Thus, new voltage is,
Vn=QCVn=62.5× 10 −612.5× 10 −6Vn=5V

Voltage of the capacitor in time t is given by,
Vn=V(1−e −t RC)5=50.0×(1−e −t 75.0×12.5× 10 −6 )1−e −t 75.0×12.5× 10 −6 =0.1e −t 75.0×12.5× 10 −6 =0.9−0.04575=−t75.0×12.5× 10 −6t=9.8775×10−5s

Energy stored in the capacitor is,

E=12CV2E=0.5×12.5×10−6×50.0×50.0E=0.01562J

Thus, new charge is,
En=0.1×0.015625En=1.5625×10−3J

So, new voltage is,
En=12CVn2Vn= 2× E n CVn=15.81V

Voltage of the capacitor in time t is given by,
Vn=V(1−e −t RC)15.81=50.0×(1−e −t 75.0×12.5× 10 −6 )1−e −t 75.0×12.5× 10 −6 =0.3162e −t 75.0×12.5× 10 −6 =0.6838−0.3800=−t75.0×12.5× 10 −6t=3.56×10−4s

Answer 7

Chapter 18, Problem 75QAP

To determine

(a)

The time required to lose 90% of its initial charge and energy

Answer 8

Expert Solution & Answer

Explanation of Solution

Given:

Capacitor, C=12.5μF=12.5×10−6F

Resistor, R=75.0Ω

Voltage of the battery, V=50.0V

Formula used:

Charge on the capacitor in time t is given by,
Q=C×V

Where,
Q= charge on the capacitor
C= capacitance
V = voltage
Energy stored in the capacitor is,
E=12CV2

Calculation:

Charge on the capacitor is given by,
Q0=C×VQ0=12.5×10−6×50.0Q0=625×10−6C

The new charge is given by.

Q=Q0−0.9Q0Q=0.1Q0Q=0.1×625×10−6Q=62.5×10−6C

Thus, new voltage is,
Vn=QCVn=62.5× 10 −612.5× 10 −6Vn=5V

Voltage of the capacitor in time t is given by,
Vn=V(1−e −t RC)5=50.0×(1−e −t 75.0×12.5× 10 −6 )1−e −t 75.0×12.5× 10 −6 =0.1e −t 75.0×12.5× 10 −6 =0.9−0.04575=−t75.0×12.5× 10 −6t=9.8775×10−5s

Energy stored in the capacitor is,

E=12CV2E=0.5×12.5×10−6×50.0×50.0E=0.01562J

Thus, new charge is,
En=0.1×0.015625En=1.5625×10−3J

So, new voltage is,
En=12CVn2Vn= 2× E n CVn=15.81V

Voltage of the capacitor in time t is given by,
Vn=V(1−e −t RC)15.81=50.0×(1−e −t 75.0×12.5× 10 −6 )1−e −t 75.0×12.5× 10 −6 =0.3162e −t 75.0×12.5× 10 −6 =0.6838−0.3800=−t75.0×12.5× 10 −6t=3.56×10−4s