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S_A_V [24]
3 years ago
10

Two equal magnitude electric charges are separated by a distance d. The electric potential at the midpoint between these two cha

rges is zero. A student considering this situation says: "The electric field at the midpoint between the two charges will be zero also, since the two charges are opposite in sign, so the fields will equal but opposite, and add to zero." There is something wrong with the student's statement. Identify any problem(s) and explain how to correct it/them.
Physics
1 answer:
Ann [662]3 years ago
8 0

Answer:

The charges under study are of the same sign

The calculation of the electric field for each charge separately, there is no relationship between the charges

Explanation:

Let's start by writing the equation for the electric field

          E = k q / r²

where q is the charge under analysis and r the distance from this charge to a positive test charge.

When analyzing the statement the student has some problems.

* The charges under study are of the same sign, it does not matter if positive or negative.

* The calculation of the electric field for each charge separately, there is no relationship between the charges for the calculation of the electric field.

* What is added is the interaction of the electric field with the positive test charge, in this case each field has the opposite direction to the other, so the vector sum gives zero

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Force F = (-6.0 N)ihat + (2.0 N) jhatacts on a particle with position vector r =(4.0 m) ihat+ (4.0 m) jhat.
finlep [7]

Answer:

\tau=(-32k)\ N-m

\theta=116.55^{\circ}

Explanation:

Given that.

Force acting on the particle, F=(-6i + 2.0j)\ N

Position of the particle, r=(4i+4j)\ m

To find,

(a) Torque on the particle about the origin.

(b) The angle between the directions of r and F

Solution,

(a) Torque acting on the particle is a scalar quantity. It is given by the cross product of force and position. It is given by :

\tau=F\times r

\tau=(-6i + 2.0j)\times (4i+4j)

\tau=\begin{pmatrix}0&0&-32\end{pmatrix}

\tau=(-32k)\ N-m

So, the torque on the particle about the origin is (32 N-m).

(b) Magnitude of r, |r|=\sqrt{4^2+4^2}=5.65\ m

Magnitude of F, |F|=\sqrt{(-6)^2+2^2}=6.324\ m

Using dot product formula,

F{\circ}\ r=|F|.|r|\ cos\theta

cos\theta=\dfrac{F{\circ} r}{|F|.|r|}

cos\theta=\dfrac{-24+8}{6.324\times 5.65}

\theta=116.55^{\circ}

Therefore, this is the required solution.

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2 years ago
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Serjik [45]

Answer:

There!!

Explanation:

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An astronomy class is so excited by the discovery of planets around other stars that they decide to do a library exhibit on the
valina [46]

Answer:

Christian Doppler

Explanation:

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3 years ago
When baking soda is mixed with lemon juice, bubbles are formed with the evolution of a gas. What type of change is it? Explain.
lorasvet [3.4K]

\large{\underline{\underline{\pmb{\frak {\color {red}{Question:}}}}}}

\sf \red{When \: baking \: soda \: is \: mixed \: with \: lemon \: juice, \: bubbles \: are \: formed \: with \: the \: evolution \: of \: a \: gas.}

\large{\underline{\underline{\pmb{\frak {\color {blue}{Answer:}}}}}}

When baking soda is mixed with lemon juice, bubbles are formed with the evolution of a gas. The gas is formed in the reaction is Carbon dioxide. CO_{2} is formed.

The change which happened in this reaction is a chemical change.

\boxed{ \frak \green{Explanation}}

Since, in chemical change we can't bring a substance to it's actual form how it was in earlier.

Examples: burning of paper is chemical, since we can't get the fine paper again after it is burnt.

Thus, the above reaction is also a chemical change, since we can't get back the lemon juice how it was earlier.

\boxed{ \frak \red{Brainlysamurai}}

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