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3 years ago

Describe how compass needle move when it is placed in a magnet ?

Physics

1 answer:

bagirrra123 [75]3 years ago

3 0

Answer:

The compass needle will point towards the magnetic field, so if it is in a magnet it will most likely spin around in circles.

Explanation:

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3 years ago

A thin, uniformly charged spherical shell has a potential of 832 V on its surface. Outside the sphere, at a radial distance of 2

Zolol [24]

Answer:

The radius is $r_1 = 0.315m$

The total charge is $Q= 2.912*10^{-8}C$

The electric potential inside sphere is the same with that outside the sphere which implies that the electric potential is 832 V

The magnitude of the electric field is $E= 2641.3 V/m$

The velocity is $v= 1.76 *10^{14} m/s$

Explanation:

From the question we are told that

The potential is $V_1 = 832 V$

The radial distance from the sphere is $d = 21.0cm = \frac{21}{100} = 0.21m$

The potential at the radial distance is $V_2 = 499V$

The potential at the surface of the sphere is mathematically represented as

$V = \frac{kQ}{r}$

$Vr = kQ$

Where kQ is a constant what this means that the the charge Q and the coulomb constant do not change

This means that

$V_1 r_1 = V_2 r_2$

Where $r_1$ is the radius of the sphere

and $r_2$ is the distance from that point where the second potential was measured to the center of the sphere which is mathematically represented as

$r_2 = r_1 + d$

Substituting this into the equation

$v_1 r_1 = V_2 (r_1 +d)$

Now substituting value

$832 * r_1 = 499 * (r_1 + 0.21)$

$832r_1 - 499r_1 = 104.79$

$333r_1 = 104.79$

$r_1 = \frac{104.79}{333}$

$r_1 = 0.315m$

From the equation above

$V = \frac{kQ}{r_1}$

making Q the subject

$Q = \frac{V r_1 }{k}$

k has a values of $k = 9*10^9 \ kg\cdot m^3 \cdot s^{-4} \cdot A^{-2}$

Substituting into the equation

$Q =\frac{832 * 0.315}{9*10^9}$

$Q= 2.912*10^{-8}C$

According to Gauss law the electric field from outside a sphere is taken to be an electric field from a point charge this mean that the potential outside a sphere is also taken as electric potential inside a sphere

The magnitude of a electric field from a sphere (point charge ) is mathematically represented as

$E = \frac{kQ}{r_1^2}$

Substituting values

$E = \frac{9*10^{9} * 2.912*10^{-8}}{0.315^2}$

$E= 2641.3 V/m$

According the the law of energy conservation

The electric potential energy at the point outside the surface where the second potential was measured(21 cm from the sphere surface) = The electric potential energy at the surface + The kinetic energy of the charge (electron) at that the surface

Generally Electric potential energy is mathematically represented as

$EPE = V * e$