How Do You Calculate the X-Component of Electrical Force on a Charge?

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SUMMARY

The x-component of the electrical force on a charge of 2.10 micro-Coulombs located at x=-3.6 m is calculated using the formula Fx = q * Ex. The electric field Ex is determined from the slope of the electric potential graph, yielding a value of 7500 N/C. The final calculation results in an x-component of the electrical force of 15.75 micro-Newtons (μN), which converts to 0.00001575 Newtons (N). Key corrections include recognizing the charge unit as micro-Coulombs and ensuring the slope is derived correctly from the potential graph.

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  • Understanding of electric potential and electric field concepts
  • Familiarity with calculus, specifically derivatives
  • Knowledge of unit conversions, particularly between micro-Coulombs and milli-Coulombs
  • Ability to interpret graphical data in physics
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  • Study the relationship between electric potential and electric field in detail
  • Learn how to derive slopes from graphs to find electric fields
  • Explore unit conversion techniques for electrical measurements
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GingerBread27
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The electric potential along the x-axis (in kv) is plotted versus the value of x, (in meters). Evaluate the x-component of the electrical force (in Newtons)on a charge of 2.10 micro-C located on the x-axis at x=-3.6 m.

Ok I 'm doing Ex=-dV/dx. I get the slope to be (-7500 N/C) from doing -(5Kv-20kV)/(0m-2m). Then I am multiplying -7500N/C by the charge of 2.10 mC to get .01575 N and I need an answer in N so I thought this was right, and it's not lol. Help?
 

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hmm... the graph isn't exactly linear in that interval, and those values don't look like they can be determined precisely just by looking at the graph ( no function is given for potential wrt x is there? ).
However checking your calculations, it looks like you multiplied -7500N/C by 2.10 micro coulombs (10^-6 C) . Micro has the symbol [itex]\mu[/itex]
but in the problem, the charge is given in milli coulombs (10^-3 C ) so your answer might be off by a factor of 10^3.
 
Last edited:


First, let's clarify a few things. The x-component of the electrical force is typically denoted as Fx, not Ex. Additionally, the slope of the electric potential graph is not the same as the electric field. The electric field is given by the negative of the derivative of electric potential with respect to distance, not position. So, we need to use the equation Fx = q * Ex, where q is the charge and Ex is the electric field.

To calculate the electric field at x=-3.6 m, we need to find the slope of the electric potential graph at that point. This can be done by finding the tangent line at x=-3.6 m. Using the slope formula, we get:

slope = (V2 - V1) / (x2 - x1) = (-7.5 kV - (-20 kV)) / (-1.8 m - (-2 m)) = 7500 N/C

Notice that we used -1.8 m instead of -3.6 m as the second x-value, since we are finding the slope at x=-3.6 m.

Now, we can calculate the x-component of the electrical force as:

Fx = (2.10 μC) * (7500 N/C) = 15.75 μN

Note that this answer is in micro-Newtons (μN), not Newtons (N). To convert to Newtons, we need to divide by 1 million, since 1 μN = 10^-6 N. So, the final answer is:

Fx = 15.75 μN / 1 million = 0.00001575 N

Therefore, the x-component of the electrical force on a charge of 2.10 μC located at x=-3.6 m is 0.00001575 N.
 

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