A parallel plate condenser has a uniform electric field \(E\)(V/m) in the space between the plates. If the distance between the plates is \(d\)(m) and area of each plate is \(A(\text{m}^2)\), the energy (joule) stored in the condenser is:
1. | \(\dfrac{1}{2}\varepsilon_0 E^2\) | 2. | \(\varepsilon_0 EAd\) |
3. | \(\dfrac{1}{2}\varepsilon_0 E^2Ad\) | 4. | \(\dfrac{E^2Ad}{\varepsilon_0}\) |
Four electric charges +q, + q, -q and -q are placed at the corners of a square of side 2L (see figure). The electric potential at point A, mid-way between the two charges +q and +q, is
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(3) Zero
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Three charges, each +q, are placed at the corners of an isosceles triangle ABC of sides BC and AC equal to 2a. D and E are the mid points of BC and CA. The work done in taking a charge Q from D to E is
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(3)zero
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A series combination of \(n_1\) capacitors, each of value \(C_1\), is charged by a source of potential difference \(4\) V. When another parallel combination of \(n_2\) capacitors, each of value \(C_2\), is charged by a source of potential difference \(V\), it has the same (total) energy stored in it as the first combination has. The value of \(C_2\) in terms of \(C_1\) is:
1. \(\frac{2C_1}{n_1n_2}\)
2. \(16\frac{n_2}{n_1}C_1\)
3. \(2\frac{n_2}{n_1}C_1\)
4. \(\frac{16C_1}{n_1n_2}\)
Three concentric spherical shells have radii a, b and c (a<b<c) and have surface charge densities and respectively. If and denote the potential of the three shells, if c=a+b, we have
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Three capacitors each of capacitance C and of breakdown voltage V are joined in series. The capacitance and breakdown voltage of the combination will be
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The electric potential at a point (x,y,z) is given by
The electric field at that point is
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The mean free path of electrons in a metal is The electric field which can give on an average 2 eV energy to an electron in the metal will be in a unit of :
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Two dielectric slabs of constant \(K_1\) and \(K_2\) have been filled in between the plates of a capacitor as shown below. What will be the capacitance of the capacitor?
1. \(\frac{2\varepsilon_0A}{2}\left(K_1+K_2\right)\)
2. \(\frac{2\varepsilon_0A}{2}\frac{\left(K_1+K_2\right)}{K_1\times K_2}\)
3. \(\frac{2\varepsilon_0A}{d}\left(\frac{K_1+K_2}{K_1-K_2}\right)\)
4. \(\frac{2\varepsilon_0A}{d}\left(\frac{K_1\times K_2}{K_1+K_2}\right)\)
What is the equivalent capacitance between A and B in the given figure (all are in farad)
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