Physics, 18.12.2019 22:31 JOHNDoe3443

Two point charges are separated by a distance of 10.0 cm. one has a charge of -25 μc and the other +50 μc. (a) determine the direction and magnitude of the electric field at a point p between the two charges that is 2.0 cm from the negative charge. (b) if an electron (mass = 9.11 x 10-31 kg) is placed at rest at p and then released, what will be its initial acceleration (direction and magnitude)?

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Answer from: wendymtz2004

6.33 x 10⁸ N/C

Direction : Towards negative charge.

1.11125 x 10²⁰ m/s²

Direction : Towards positive charge.

Explanation:

$Q_{1}$ = magnitude of negative charge = 25 x 10⁻⁶ C

$Q_{2}$ = magnitude of positive charge = 50 x 10⁻⁶ C

$r_{1}$ = distance of negative charge from point P = 0.02 m

$r_{2}$ = distance of positive charge from point P = 0.08 m

Magnitude of electric field at P due to negative charge is given as

$E_{1} = \frac{kQ_{1}}{r_{1}^{2} } = \frac{(9\times10^{9})(25\times10^{-6})}{0.02^{2} } = 5.625\times10^{8} N/C$

Magnitude of electric field at P due to positive charge is given as

$E_{2} = \frac{kQ_{2}}{r_{2}^{2} } = \frac{(9\times10^{9})(50\times10^{-6})}{0.08^{2} } = 0.703125\times10^{8} N/C$

Net electric field at P is given as

$E = E_{1} + E_{2}\\E = 5.625\times10^{8} + 0.703125\times10^{8} \\E = 6.33\times10^{8} N/C$

Direction:

Towards the negative charge.

= mass of the electron placed at P = 9.31 x 10⁻³¹ C

$Q_{1}$ = magnitude of charge on electron = 1.6 x 10⁻¹⁹ C

Acceleration of the electron due to the electric field at P is given as

$a = \frac{qE}{m}\\ a = \frac{(1.6\times10^{-19})(6.33\times10^{8})}{(9.11\times10^{-31})}\\a = 1.11125\times10^{20} ms^{-2}$

Direction: Towards the positive charge Since a negative charge experience electric force in opposite direction of the electric field.

Two point charges are separated by a distance of 10.0 cm. one has a charge of -25 μc and the other +

Answer from: major126

Explanation:

qA = - 25 x 10^-6 C

qB = 50 x 10^-6 C

AP = 2 cm

BP = 8 cm

(a)

Electric field at P due to the charge at A

$E_{A}=\frac{kq_{A}}{AP^{2}}$

$E_{A}=\frac{9\times 10^{9}\times 25\times 10^{-6}}{0.02^{2}}$

EA = 5.625 x 10^8 N/C

Electric field at P due to the charge at B

$E_{B}=\frac{kq_{B}}{BP^{2}}$

$E_{A}=\frac{9\times 10^{9}\times 50\times 10^{-6}}{0.08^{2}}$

EB = 0.70 x 10^8 N/C

The resultant electric field at P due to both the charges is

E = EA+ EB = (5.625 + 0.7) x 10^8

E = 6.325 x 10^8 N/C towards left

(b) mass of electron, m = 9.1 x 10^-31 kg

Let a be the acceleration of electron

Force on electron, F = charge of electron x electric field

F = q x E

$a = \frac{qE}{m}$

$a = \frac{1.6\times 10^{-19}\times 6.325\times 10^{8}}{9.1\times 10^{-31}}$

a = 1.11 x 10^20 m/s^2

Answer from: Quest

Then the answer would be d then

Answer from: Quest

we have been given a quadratic function $f(x)=(x-5)^{2} +2$ and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is $[2,\infty)$ .

if we restrict the domain of this function to either $(-\infty,5]$ or $[5,\infty)$ , it will become one to one function.

let us know find its inverse.

$y=(x-5)^{2}+2$

upon interchanging x and y, we get:

$x=(y-5)^{2}+2$

let us now solve this function for y.

$(y-5)^{2}=x-2\\ y-5=\pm \sqrt{x-2}\\ y=5\pm \sqrt{x-2}\\$

hence, the inverse function would be $f^{-1}(x)=5+\sqrt{x-2}$ if we restrict the domain of original function to $[5,\infty)$ and the inverse function would be $f^{-1}(x)=5-\sqrt{x-2}$ if we restrict the domain to $(-\infty,5]$ .

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Two point charges are separated by a distance of 10.0 cm. one has a charge of -25 μc and the other +...

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