Factors Affecting Capacitance in a Capacitor

  • Thread starter Thread starter Gear2d
  • Start date Start date
  • Tags Tags
    Capacitance
AI Thread Summary
Capacitance in a capacitor can be increased by decreasing the distance between the plates, increasing the charge, decreasing the electric field, and increasing the area of the plates, while thickness does not affect capacitance. The equation C = εA/d is more relevant for understanding these factors than Q = CV. A stronger electric field can attract more charges, thus increasing capacitance. Additionally, increasing the dielectric constant (k) will also enhance capacitance. Understanding these principles is crucial for analyzing capacitor behavior in electrical circuits.
Gear2d
Messages
49
Reaction score
0

Homework Statement



What will increase the capacitance of a capacitor?

Homework Equations



Q=CV
V = Ed

The Attempt at a Solution



Factors that increase:
-Decrease in distance
-Increase in charge
-Decrease in electric field
-Increase in area (length and width of the capacitor, but its thickness has no affect)

Would these be correct, also what about the dielectric constant (k), would increasing it increase capacitance or decrease it?
 
Physics news on Phys.org


Gear2d said:

Homework Statement



What will increase the capacitance of a capacitor?

Homework Equations



Q=CV
V = Ed

The Attempt at a Solution



Factors that increase:
-Decrease in distance
-Increase in charge
-Decrease in electric field
-Increase in area (length and width of the capacitor, but its thickness has no affect)

Would these be correct, also what about the dielectric constant (k), would increasing it increase capacitance or decrease it?

Q = VC is not exactly the right equation to consider. The capacitor is the constant that relates the Charge to Voltage and is not a property of the capacitor per se. If you change the voltage then the charge changes as determined by the C - it is just descriptive of the relationship between Q and V as defined by C..

On the other hand:

C = εA/d would be more to the point of the question.

For more on capacitors read the link - https://www.physicsforums.com/library.php?do=view_item&itemid=112
 


Hello Gear2d, I am Gear300...
Its pretty much what LowlyPion said. C = Q/V is somewhat deceptive. If you think about it, a stronger electric field should bring in more charges...and it does. If you decrease the value of E, then you decrease the value of Q. C is the constant, and what you would want is a more stable equation for it (provided by LowlyPion). If you want to further adjust the equation, then C = εA/d = Q/V = Q/(Ed)...Q = εAE...so charge increases with increasing electric field. Increasing the dielectric constant k increases the capacitance. If you want to know how, look for something correlating with a microscopic observation of capacitance and dielectrics.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top