The transformer is based on:

A tiny object carrying a charge of +44 μC and a second tiny charged object are initially very far apart. If it takes 21 J of wor

STatiana [176]

Answer:

The magnitude of the second charge is $\rm 1.062\times 10^{-7}\ C$ or $\rm 0.1062\ \mu C.$

Explanation:

The work done in bringing a charged particle from one point to another in the presence of some electric field is equal to the change in the electric potential energy of the charge in moving from one point to another.

The electric potential energy of some charge $q_o$ at a point in the electric field of another charge $q$ is given by the product of the amount of charge $q_o$ and electric potential at that point due to the charge $q$ .

$U = q_o\ V.$

The electric potential at that point is given by

$V = \dfrac{kq}{r}.$

where $k$ is the Coulomb's constant.

Therefore,

$U=q_o\ \dfrac{kq}{r}.$

Now, We have given two charges $q_1 = +44\ \mu C = +44\times 10^{-6}\ C$ and $q_2$ , whose value is to be found.

When the two charges are infinitely dar apart, the electric potential energy of the system is given by

$U_i = \dfrac{kq_1q_2}{\infty}=0.$

When the coordinates of position of the two charges are

$(x_1,\ y_1) = (1.00\ mm,\ 1.00\ mm).\\(x_2,\ y_2) = (1.00\ mm,\ 3.00\ mm).$

The distance between the two charges is given by

$r=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}=\sqrt{(1.00-1.00)^2+(3.00-1.00)^2}=2.00\ mm = 2.00\times 10^{-3}\ m.$

The electric potential energy of the charges in this configuration is given by

$U_f = \dfrac{kq_1q_2}{r}\\=\dfrac{(8.99\times 10^9)\times (+44\times 10^{-6})\times q_2}{2.00\times 10^{-3}}\\=1.9778\times 10^8\times q_2.$

The change in the electric potential energy of the system is equal to the work done to bring the system from inifinitely far apart position to given configuration.

Therefore,

$W = U_f-U_i\\21=(1.9778\times 10^8\times q_2)-0\\\Rightarrow q_2 = \dfrac{21}{1.9778\times 10^8}\\=1.062\times 10^{-7}\ C\\=0.1062\times 10^{-6}\ C\\=0.1062\ \mu C.$

6 0

3 years ago

____________ stays the same no matter what location you are in.

alexdok [17]

Answer:

the date is the only thing that's the same everywhere you go because time is different

6 0

3 years ago

Hypothesis:

posledela

They are formed when two plates collide, either crumpling up and forming mountains or pushing one of the plates under the other and back into the mantle to melt. Convergent boundaries form strong earthquakes, as well as volcanic mountains or islands, when the sinking oceanic plate melts.

5 0

3 years ago

A surface will be an equipotential surface if (there may be more than one correct choice).A. the electric field is zero at all p

aev [14]

Answer:

c. A, C

Explanation:

On equi-potential surface, fields are equal in magnitude at all points . If field is zero at all points , they will be equal so first option is correct.

Field is perpendicular to equipotential surface at all points.

6 0

3 years ago

In an experiment designed to measure the strength of a uniform magnetic field produced by a set of coils, electrons are accelera

Arlecino [84]

Answer:

the magnitude of the magnetic field is 8.704 x 10⁻⁴ T

Explanation:

Given;

potential difference, V = 278 V

radius of the circular path, r = 6.46 cm = 0.0646 m

charge of electron, q = 1.60218 × 10⁻¹⁹ C

mass of electron, m = 9.10939 × 10⁻³¹ kg

The magnitude of the magnetic field is given as;

$B = \frac{M_e*v}{q*r}$

where;

B is the magnitude of the magnetic field

$M_e$ is mass of the electron

v is velocity of the electron

r is the radius of the circular path

q is charge of the electron

Determine velocity of the electron from kinetic energy equation;

$K = \frac{1}{2} M_ev^2\\\\Vq = \frac{1}{2} M_ev^2\\\\v^2 = \frac{2qV}{M_e} \\\\v = \sqrt{\frac{2qV}{M_e}} = \sqrt{\frac{2*1.602*10^{-19}*278}{9.109*10^{-31}}} = 9.8886*10^{6} \ m/s$

the magnitude of the magnetic field:

$B = \frac{M_e*v}{q*r} \\\\B = \frac{(9.109*10^{-31})*(9.8886*10^6)}{(1.602*10^{-19})*(0.0646)} = 8.704*10^{-4} \ T$

Therefore, the magnitude of the magnetic field is 8.704 x 10⁻⁴ T

5 0

3 years ago