Let us situate this on the x axis, and let our uniform line of charge be positioned on the interval <span>(−L,0]</span> for some large number L. The voltage V as a function of x on the interval <span>(0,∞)</span> is given by integrating the contributions from each bit of charge. Let the charge density be λ. Thus, for an infinitesimal length element <span>d<span>x′</span></span>, we have <span>λ=<span><span>dq</span><span>d<span>x′</span></span></span></span>.<span>V(x)=<span>1/<span>4π<span>ϵ0</span></span></span><span>∫line</span><span><span>dq/</span>r</span>=<span>λ/<span>4π<span>ϵ0</span></span></span><span>∫<span>−L</span>0</span><span><span>d<span>x/</span></span><span>x−<span>x′</span></span></span>=<span>λ/<span>4π<span>ϵ0</span></span></span><span>(ln|x+L|−ln|x|)</span></span>
Answer:
E = 1.04*10⁻¹ N/C
Explanation:
Assuming no other forces acting on the proton than the electric field, as this is uniform, we can calculate the acceleration of the proton, with the following kinematic equation:
As the proton is coming at rest after travelling 0.200 m to the right, vf = 0, and x = 0.200 m.
Replacing this values in the equation above, we can solve for a, as follows:
According to Newton´s 2nd Law, and applying the definition of an electric field, we can say the following:
F = mp*a = q*E
For a proton, we have the following values:
mp = 1.67*10⁻²⁷ kg
q = e = 1.6*10⁻¹⁹ C
So, we can solve for E (in magnitude) , as follows:
⇒ E = 1.04*10⁻¹ N/C
Yes it is, it was made in France as a gift to the U.S.A.
Explanation:
To find the resultant force subtract the magnitude of the smaller force from the magnitude of the larger force. The direction of the resultant force is in the same direction as the larger force
It takes about a really long time
