Given k=100, Area if plate A = 1.00 cm2

Thickness of rutile 

Without any dielectric medium capacitance of a parallel plate capacitor is c = oAd
When a dielectric is placed between capacitors plates then capacitance c1 = KoAd
Here
, 
Now 



Correct Option: A
What purposes serve parallel plate capacitors?
The following are some uses for parallel plate capacitors:
- This kind of capacitor is utilized in rechargeable energy systems such as batteries.
- These capacitors are used in systems for dynamic digital memory.
- Such capacitors are used in pulsed laser and radar circuits.
To learn more about Parallel plate capacitors, visit:
brainly.com/question/12733413
#SPJ4
The tension in the string supporting the rod and the attached weights is 7.11 N.
<h3>Tension in the string supporting the rod</h3>
The tension in the string supporting the rod and the attached weights is the sum of the weights supported by the strings.
T = (m1 + m2)g
where;
- m1 and m2 are the two masses supported
0.4lb + 1.2lb = 1.6lb = 0.725 kg
T = 0.725 x 9.8
T = 7.11 N
Learn more about tension here: brainly.com/question/918617
#SPJ1
0.1 J of elastic potential energy will be stored by the spring when it gets compressed by 0.2 m if it has the spring constant of 5 N/m.
Elastic potential energy can be defined as the potential energy stored by the spring when it is stretched or compressed.
- Amount of energy stored in the elastic material or spring is directly proportional to the amount of compression or elongation.
- It is represented by U.
- Mathematically, U =
; k is the spring constant and x is the compression or expansion suffered by the spring
According to the question.
Compression in the spring, x = 0.2 m
Spring constant, k = 5 n/m
Elastic potential Energy, U = 
U = 0.1 J
0.1 J of elastic potential energy will be stored by the spring.
To know more about elastic potential energy,
brainly.com/question/2611925
#SPJ1