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# Consider the charges $$q,~q,$$ and $$-q$$ placed at the vertices of an equilateral triangle, as shown in the figure. Then the sum of the forces on the three charges is:      1. $$\frac{1}{4\pi \epsilon _{0}}\frac{q^{2}}{l^{2}}$$ 2. zero 3. $$\frac{2}{4\pi \epsilon _{0}}\frac{q^{2}}{l^{2}}$$ 4. $$\frac{3}{4\pi \epsilon _{0}}\frac{q^{2}}{l^{2}}$$ 

Subtopic:  Coulomb's Law |
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An electron falls through a distance of $$1.5$$ cm in a uniform electric field of magnitude $$2\times10^4$$ N/C [figure (a)]. The direction of the field is reversed keeping its magnitude unchanged and a proton falls through the same distance [figure (b)]. If $$t_e$$ and $$t_p$$ are the time of fall for electron and proton respectively, then:

1. $$t_e=t_p$$
2. $$t_e>t_p$$
3. $$t_e<t_p$$
4. none of these


Subtopic:  Electric Field |
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Two-point charges ${q}_{1}$ and ${q}_{2}$, of magnitude   and , respectively, are placed 0.1 m apart. The electric field at point  A (as shown in the figure) is:

Subtopic:  Electric Field |
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Two charges $$\pm10~\mu\text{C}$$ are placed $$5.0$$ mm apart. The electric field at a point $$P$$ on the axis of the dipole $$15$$ cm away from its centre $$O$$ on the side of the positive charge, as shown in the figure is:

1. $$2.7\times10^5~\text{NC}^{-1}$$
2.
$$4.13\times10^6~\text{NC}^{-1}$$
3. $$3.86\times10^6~\text{NC}^{-1}$$
4. $$1.33\times10^5~\text{NC}^{-1}$$

Subtopic:  Electric Dipole |
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Two charges $$\pm 10~ \mu \text{C}$$ are placed $$5.0$$ mm apart. The electric field at a point $$Q$$, $$15$$ cm away from O on a line passing through $$O$$ and normal to the axis of the dipole, as shown in the figure is:

1. $$2.8 \times 10^5~\text{NC}^{-1}$$
2. $$3.9\times 10^5 ~\text{NC}^{-1}$$
3. $$1.33\times 10^5 ~\text{NC}^{-1}$$
4. $$4.1\times 10^6 ~\text{NC}^{-1}$$


Subtopic:  Electric Dipole |
68%
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The electric field components in the shown figure are ${\mathrm{E}}_{\mathrm{x}}=\alpha {x}^{1/2}$Ey = Ez = 0, in which  . The net flux through the cube is: (Assume that a = 0.1 m)

Subtopic:  Gauss's Law |
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The electric field components in the shown figure are ${\mathrm{E}}_{\mathrm{x}}={\mathrm{ax}}^{1/2}$Ey = Ez = 0, in which  . Find the charge within the cube if net flux through the cube is : (Assume that a = 0.1 m).

Subtopic:  Gauss's Law |
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An electric field is uniform, and in the positive $$x$$-direction for positive $$x$$, and uniform with the same magnitude but in the negative $$x$$-direction for negative $$x$$. It is given that $$\vec{E}=200\hat{i}$$ N/C for $$x>0$$ and $$\vec{E}=-200\hat{i}$$  N/C for $$x<0$$. A right circular cylinder of length $$20~\text{cm}$$ and radius $$5~\text{cm}$$ has its centre at the origin and its axis along the $$x\text{-}$$axis so that one face is at $$x= + 10~\text{cm}$$ and the other is at $$x= -10~\text{cm}$$ (as shown in the figure). What is the net outward flux through the cylinder?

1. $$0$$
2. $$1.57~\text{Nm}^2\text{C}^{-1}$$
3. $$3.14~\text{Nm}^2\text{C}^{-1}$$
4. $$2.47~\text{Nm}^2\text{C}^{-1}$$

Subtopic:  Gauss's Law |
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An electric field is uniform, and in the positive x-direction for positive x, and uniform with the same magnitude but in the negative x-direction for negative x. It is given that E = N/C for x > 0 and E = – N/C for x < 0. A right circular cylinder of length 20 cm and radius 5 cm has its centre at the origin and its axis along the x-axis so that one face is at x = +10 cm and the other is along the x-axis so that one face is at x = +10 cm and the other is at x = –10 cm (as shown in the figure). What is the net charge inside the cylinder?

Subtopic:  Gauss's Law |
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An early model for an atom considered it to have a positively charged point nucleus of charge Ze, surrounded by a uniform density of negative charge up to a radius R. The atom as a whole is neutral. For this model, the electric field at a distance r(r<R) from the nucleus is:

Subtopic:  Gauss's Law |
57%
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