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Bezzdna [24]
3 years ago
7

Assume that the electric field E is equal to zero at a given point. Does it mean that the electric potential V must also be equa

l to zero at this point? If not, then provide an example to prove your answer. If you think that the answer is "yes", please justify it.
Physics
1 answer:
lyudmila [28]3 years ago
8 0

Answer:

  • No, this doesn't mean the electric potential equals zero.

Explanation:

In electrostatics, the electric field \vec{E} is related to the gradient of the electric potential V with :

\vec{E} (\vec{r}) = - \vec{\nabla} V (\vec{r})

This means that for constant electric potential the electric field must be zero:

V(\vec{r}) = k

\vec{E} (\vec{r}) = - \vec{\nabla} V (\vec{r}) = - \vec{\nabla} k

\vec{E} (\vec{r}) = -  (\frac{\partial}{\partial x} , \frac{\partial}{\partial y } , \frac{\partial}{\partial z}) k

\vec{E} (\vec{r}) = -  (\frac{\partial k}{\partial x} , \frac{\partial k}{\partial y } , \frac{\partial k}{\partial z})

\vec{E} (\vec{r}) = -  (0,0,0)

This is not the only case in which we would find an zero electric field, as, any scalar field with gradient zero will give an zero electric field. For example:

V(\vec{r})= (x+2)^2 (y+4)^3 (z+5)^4

give an electric field of zero at point (0,0,0)

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Explanation:

1)

The motion of the bullet is a projectile motion, which consists of two separate motions:  

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From the equation of motions along the x- and y- directions, it is possible to find an expression for the horizontal range covered by a projectile, and it is:

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Solving the equation, we find the horizontal range:

d=\frac{(100)^2sin(2\cdot 60^{\circ})}{9.8}=884 m

2)

To find the maximum height, we have to analyze the vertical motion of the bullet. We can do it by using the following suvat equation:

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v_y is the vertical velocity of the bullet after having covered a vertical displacement of s

u_y is the initial vertical velocity

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The vertical component of the initial velocity is given by

u_y = u sin\theta

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Substituting into the equation and re-arranging for s, we find the maximum height:

s=\frac{u^2 sin^2 \theta}{2g}=\frac{(100)^2(sin 60^{\circ})^2}{2(9.8)}=383 m

Learn more about projectile motion:

brainly.com/question/8751410

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