Physics, 19.07.2019 00:20 martdrea13

Along the line connecting the two charges, at what distance from the charge q1 is the total electric field from the two charges zero? express your answer in terms of some or all of the variables s, q1, q2 and k =14πϵ0

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Answer from: jayzeptor

Distance of zero electric field from charge q1 = $\frac{a\sqrt{q1}}{(\sqrt{{q2}}+\sqrt{q1})}$

Explanation:

let the distance between the charges be 'a'

and the point of equal electric field be at a distance 'x' from the charge q1

Given:

the charges as q1 and q2

we know that the Electric field (E) is given as

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

where,

k = Electric field constant

Q = charge

r = distance of the point from the charge where Electric field is being measures

⇒let the distance between the charges be 'a'

and the point of equal electric field be at a distance 'x' from the charge q1

⇒Now electric field due to the charges q1 = $E_1=\frac{kq1}{r_1^2}$

electric field due to the charges q2 = $E_1=\frac{kq2}{r_2^2}$

from the figure attached: r_1=x and r_2=a-x

for the electric field from the charges to be zero

$\frac{kq1}{r_1^2}=\frac{kq2}{r_2^2}$

substituting the values

$\frac{kq1}{a^2}=\frac{kq2}{(a-x)^2}$

$\frac{a-x}{x}=\sqrt{\frac{q2}{q1}}$

$\frac{a}{x}-\frac{x}{x}=\sqrt{\frac{q2}{q1}}$

$a=(\sqrt{\frac{q2}{q1}}+1)x$

$x = \frac{a}{(\sqrt{\frac{q2}{q1}}+1)}$

$x = \frac{a\sqrt{q1}}{(\sqrt{{q2}}+\sqrt{q1})}$

Answer from: Quest

the velocity you are looking for is made up of two parts: horizontally velocity, vh, and its vertical velocity, vv. the actual velocity of the ball will be these two parts added together, and because velocity is a vector then they will need to be added vectorially.

neglecting air resistance etc etc.,

the horizontal velocity component will be constant (because there is no accelerating force in the horizontal direction.

vh = 8 ms-1

the vertical velocity component has an accelerating force: 'g'.

using v = u +at where a = g = 9.81 ms-2.

the vertical component of velocity, vv = 0 + gt = vv =9.81 x 2

vv = 19.62 ms-1 . the zero appears because the virtical component of velocity at time t=0 is zero. remember, the ball was thrown horizontally.

how fast it will be moving (it actual path (being a parabola )) will be the vector sum of these two.

adding vectorally gives: -

v = sqr root (8^2 + 19.62^2) = sqr root(64+384.9) = 21.18 or 21.2 ms-1

Answer from: Quest

the only real purpose of sputnik 1 was to beat the us into space. we were getting close to launching a useful satellite, explorer 1 - which launched on january 31, 1958. the ussr had to beat that, so they launched sputnik 1 on october 4, 19

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