Question:
A capacitor has capacitance \(C_0\) when there is no dielectric between its plates. Two slabs of dielectric constants \(K_1\) and \(K_2\), each having area equal to the area of the plates but thickness equal to half of the separation between the plates, are placed between the plates. Then the new capacitance is:
A capacitor has capacitance \(C_0\) when there is no dielectric between its plates. Two slabs of dielectric constants \(K_1\) and \(K_2\), each having area equal to the area of the plates but thickness equal to half of the separation between the plates, are placed between the plates. Then the new capacitance is:
Show Hint
When dielectric slabs are placed one after another along the electric field direction, they behave like capacitors connected in series. Always find the individual capacitances first and then apply the series combination formula.
Updated On: Jun 26, 2026
- \(C_0(K_1+K_2)\)
- \(C_0\left(\dfrac{K_1K_2}{K_1+K_2}\right)\)
- \(C_0\left(\dfrac{K_1+K_2}{K_1K_2}\right)\)
- \(2C_0\left(\dfrac{K_1K_2}{K_1+K_2}\right)\)
Show Solution
Verified By Collegedunia
The Correct Option is D
Solution and Explanation
Step 1: Determine the arrangement of dielectric slabs.
The two dielectric slabs have area equal to the plate area and each occupies half of the plate separation.
Hence, the slabs are placed one after another along the electric field direction. Therefore, they behave as two capacitors connected in series.
Let the plate separation be \[ d \] Then each slab has thickness \[ \frac{d}{2} \] and plate area \[ A \]
Step 2: Find the capacitance of the first dielectric slab.
For dielectric constant \(K_1\), \[ C_1=\frac{K_1\varepsilon_0A}{d/2} \] \[ C_1=\frac{2K_1\varepsilon_0A}{d} \] Since \[ C_0=\frac{\varepsilon_0A}{d} \] therefore, \[ C_1=2K_1C_0 \]
Step 3: Find the capacitance of the second dielectric slab.
Similarly, for dielectric constant \(K_2\), \[ C_2=\frac{K_2\varepsilon_0A}{d/2} \] \[ C_2=\frac{2K_2\varepsilon_0A}{d} \] \[ C_2=2K_2C_0 \]
Step 4: Use the series combination formula.
Since the two dielectric regions act as capacitors in series, \[ \frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} \] Substituting \(C_1\) and \(C_2\), \[ \frac{1}{C} = \frac{1}{2K_1C_0} + \frac{1}{2K_2C_0} \] \[ \frac{1}{C} = \frac{1}{2C_0} \left( \frac{1}{K_1} + \frac{1}{K_2} \right) \] \[ \frac{1}{C} = \frac{K_1+K_2} {2C_0K_1K_2} \] Therefore, \[ C= \frac{2C_0K_1K_2} {K_1+K_2} \]
Step 5: Final conclusion.
Hence, the new capacitance is \[ \boxed{ 2C_0\left(\frac{K_1K_2}{K_1+K_2}\right) } \]
The two dielectric slabs have area equal to the plate area and each occupies half of the plate separation.
Hence, the slabs are placed one after another along the electric field direction. Therefore, they behave as two capacitors connected in series.
Let the plate separation be \[ d \] Then each slab has thickness \[ \frac{d}{2} \] and plate area \[ A \]
Step 2: Find the capacitance of the first dielectric slab.
For dielectric constant \(K_1\), \[ C_1=\frac{K_1\varepsilon_0A}{d/2} \] \[ C_1=\frac{2K_1\varepsilon_0A}{d} \] Since \[ C_0=\frac{\varepsilon_0A}{d} \] therefore, \[ C_1=2K_1C_0 \]
Step 3: Find the capacitance of the second dielectric slab.
Similarly, for dielectric constant \(K_2\), \[ C_2=\frac{K_2\varepsilon_0A}{d/2} \] \[ C_2=\frac{2K_2\varepsilon_0A}{d} \] \[ C_2=2K_2C_0 \]
Step 4: Use the series combination formula.
Since the two dielectric regions act as capacitors in series, \[ \frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} \] Substituting \(C_1\) and \(C_2\), \[ \frac{1}{C} = \frac{1}{2K_1C_0} + \frac{1}{2K_2C_0} \] \[ \frac{1}{C} = \frac{1}{2C_0} \left( \frac{1}{K_1} + \frac{1}{K_2} \right) \] \[ \frac{1}{C} = \frac{K_1+K_2} {2C_0K_1K_2} \] Therefore, \[ C= \frac{2C_0K_1K_2} {K_1+K_2} \]
Step 5: Final conclusion.
Hence, the new capacitance is \[ \boxed{ 2C_0\left(\frac{K_1K_2}{K_1+K_2}\right) } \]
Was this answer helpful?
0
0
Top AP EAPCET Physics Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The radii of the two columns in a U - tube are r1 and r2. When a liquid of density p (angle of contact is $ 0{}^\circ $ ) is filled in it, the level difference of the liquid in the two arms is h. The surface tension of the liquid is: (g = acceleration due to gravity)
- A rigid metallic sphere is spinning around its own axis in the absence of external torque. If the temperature is raised, its volume increases by $9 \%$. The change in its angular speed is
View More Questions
Top AP EAPCET Capacitance Questions
- When a parallel plate capacitor is charged up to 95 V, its capacitance is \( C \). If a dielectric slab of thickness 2 mm is inserted between plates and the plate separation is increased by 1.6 mm such that the potential difference remains constant, find the dielectric constant of the material:
Four metal plates, each with surfa area $A$ on one side, are plad with separation $d$ as shown in the figure. The capacitan between a and b is ($\varepsilon_0$ -- permittivity of free spa)
- Between the plates of a parallel plate capacitor of plate area A and capacity 0.025 \( \mu \)F, a metal plate of area A and thickness equal to \( \frac{1}{3} \) of the separation between the plates of the capacitor is introduced. If the capacitor is charged to 100 V, then the amount of work done to remove the metal plate from the capacitor is
- A capacitor of capacitance 2 μF is charged with the help of a 60 V battery. After disconnecting the battery, if this capacitor is connected in parallel with another uncharged capacitor of capacitance 1 μF, then the potential difference across the plates of the 2 μF capacitor is
- In the given circuit, the internal resistance of the cell is zero. If \( i_1 \) and \( i_2 \) are the readings of the ammeter when the key (K) is opened and closed respectively, then \( i_1 : i_2 = \)
View More Questions
Top AP EAPCET Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- The four quantum numbers of the valence electron of potassium are :
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
View More Questions