What value must q2 have if the electric potential at point a is to be zero?

1 answer:

5 0

<span>let r be side of square
V (B) = k q2/r + k Q / sqrt[r^2 + r^2] = 0 given
q2/r + Q / r root2 = 0
q2 = - Q / root2
q2/Q = - 1/root2
q2/Q = - root2 /2 = - 0.707

q2 equals negative 0.707 times Q</span>

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A proton is accelerated down a uniform electric field of 450 N/C. Calculate the acceleration of this proton.

olasank [31]

Answer:

The acceleration of proton will be $431.137\times 10^8m/sec^2$

Explanation:

We have given electric field E = 450 N/C

Charge on proton $=1.6\times 10^{-19}C$

Force on electron due to electric field is given by $F=qE=1.6\times 10^{-19}\times 450=720\times 10^{-19}N$

Mass of electron $m=1.67\times 10^{-27}kg$

Now according to second law of motion $F=ma$

So $720\times10^{-19}=1.67\times 10^{-27}a$

$a=431.137\times 10^8m/sec^2$

So the acceleration of proton will be $431.137\times 10^8m/sec^2$

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luda_lava [24]

Answer:

The direction will be $84.86^\circ$ and the distance 250.75km.

Explanation:

Let's say A is the displacement vector which represents the first 170km and B the one for the next 230km. Then the components of these vector will be:

$A_x=170cos(68^{\circ})\\ A_y=170sin(68^\circ)\\\\B_x=230cos(-36^\circ)\\B_y=230sin(-36^\circ)$

The vector which point from the origin to the final position of the plane will be R=A+B. We sum components on <em>x </em>and <em>y </em>independetly (vector property):

$R_x=A_x+B_x=170cos(68^{\circ})+230cos(-36^\circ)=63.68km+186.07km=249.75km$