**Q.1) Two capacitor made of square plates of edge ‘a’ with resistor & a battery of V volts is connected in series. Separations between plates of 1^st capacitor is ‘d₁ ’ & that of 2^nd one is ‘d₂’ at steady state, an electron is projected between the plates of lower capacitor from its lower plates. What minimum speed can the electron be projected that it does not collide with any plate?

*Q.2) A capacitor of capacitance C is charged by q₀ .At t=0, it is connected to a battery of emf V and internal resistance r. find the charge on the capacitor at a time‘t’.

(getting two answers when solving taking cases as (I) connecting (+) terminals together and (II) (+) of capacitor & (-) terminal of battery but answer key says only on answer is true for both cases)q.16 of capacitance sheet

**Q.3) A parallel plate capacitor is located horizontally so that one of its plates is over its surface, the permittivity of the liquid is equal to e ,its density is s. to what height will the level of the liquid in the capacitor rise after its plates get a charge of surface density x . (Seen in physics for you from mtg publishers)

****Q.4) A current carrying wire is placed in the grooves of an insulating semicircular disc CABD of radius r. A &B are points on the disc such that current enters at point A & leaves at point B. determine the magnetic field at D( here CD is the diameter of disc also LACB = 30⁰ & LBCD= 30⁰ ) { asked in REE-93 as given in my friends modules}

**Q.5) A square loop of wire of edge ‘a’ carries current I .

a) Find the magnetic induction B for the point on the axis of the loop at a distance x from its centre.

b) Does square loop behave as dipole for points such that x >> a. If so ,what is its dipole moment?

(asked in MTG PFY)

Let's tackle your questions one by one, starting with the first one about the two capacitors connected in series. This involves some concepts from electrostatics and motion of charged particles. I’ll break it down step by step.

Understanding the Electron's Motion Between Capacitor Plates

In your scenario, we have two capacitors connected in series, and we want to find the minimum speed at which an electron can be projected between the plates of the lower capacitor without colliding with them. To do this, we need to consider the electric field created by the capacitor and the forces acting on the electron.

Electric Field Between the Plates

The electric field (E) between the plates of a capacitor can be calculated using the formula:

• E = V/d

where V is the voltage across the capacitor and d is the separation between the plates. For the lower capacitor with separation d₁, the electric field will be:

• E₁ = V₁/d₁

Force on the Electron

The force (F) acting on the electron due to the electric field is given by:

• F = eE

where e is the charge of the electron (approximately 1.6 x 10^-19 C). Thus, the force on the electron becomes:

• F = e(V₁/d₁)

Motion of the Electron

When the electron is projected with an initial speed (v₀), it will experience a downward acceleration due to the electric field. The equation of motion can be described as:

• y(t) = v₀t + (1/2)(-a)t²

Here, 'a' is the acceleration due to the electric field, which can be expressed as:

• a = F/m = eE/m = e(V₁/d₁)/m

To ensure that the electron does not collide with the plates, it must travel the distance d₁ before the downward displacement due to the electric field causes it to hit the plate. Setting the displacement equal to d₁ gives us:

• d₁ = v₀(d₁/v₀) + (1/2)(-e(V₁/d₁)/m)(d₁/v₀)²

Solving this equation will yield the minimum speed v₀ required for the electron to avoid collision.

Charge on the Capacitor Over Time

Now, let’s move on to your second question regarding the charge on a capacitor connected to a battery with internal resistance. This involves understanding how charge accumulates over time in an RC circuit.

Charging a Capacitor

When a capacitor of capacitance C is connected to a battery of emf V and internal resistance r, the charge (q) on the capacitor at time t can be described by the equation:

• q(t) = C * V * (1 - e^-t/(RC))

This equation shows that the charge on the capacitor increases exponentially over time, approaching the maximum charge of CV as time goes to infinity.

Two Cases of Connection

In your question, you mentioned two cases of connecting the capacitor to the battery:

• Case I: Connecting the positive terminal of the battery to the positive plate of the capacitor.
• Case II: Connecting the positive terminal of the battery to the negative plate of the capacitor.

In Case I, the charge will build up positively on the capacitor, while in Case II, the capacitor will charge negatively. However, the magnitude of the charge will still follow the same exponential growth pattern, leading to the same final charge magnitude, but with opposite signs.

Liquid Level Rise in a Capacitor

Next, let’s discuss the parallel plate capacitor submerged in a liquid. When the plates are charged, they create an electric field that can affect the liquid level.

Height of Liquid Rise

The electric field between the plates exerts a force on the liquid, causing it to rise. The relationship can be derived from the balance of forces acting on the liquid:

• Pressure due to liquid height: P = ρgh
• Electric pressure due to surface charge density: P = x/ε

Setting these equal gives us:

• ρgh = x/ε

From this, we can solve for the height (h) of the liquid:

• h = (x/(ρgε))

Magnetic Field in a Semicircular Disc

Now, let’s analyze the magnetic field at point D due to a current-carrying wire placed in a semicircular disc. This involves applying the Biot-Savart law.

Magnetic Field Calculation

The magnetic field (B) at a point due to a current-carrying wire can be calculated using:

• B = (μ₀I)/(4πr) * sin(θ)

In your case, we need to consider the contributions from both segments of the wire and their respective angles. The symmetry of the semicircular disc simplifies the calculations, allowing us to find the resultant magnetic field at point D.

Magnetic Induction from a Square Loop

Lastly, let’s explore the magnetic induction from a square loop of wire carrying current. This involves calculating the magnetic field at a point along the axis of the loop.

Magnetic Field at a Distance

The magnetic induction (B) at a point on the axis of a square loop can be derived using the Bi