Calculate the work needed to move a test charge from point B to D

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To calculate the work needed to move a 1.5μC test charge from point B to D, the formula W = q(Va - Vb) is appropriate, where Va and Vb represent the electric potentials at points A and B, respectively. The electric field's uniformity is not a requirement here, as the work can be related to electric potential differences instead. The potential at point B must be determined considering the influence of the nearby 75μC charges, while the potential at point D is calculated similarly. The distances used in the calculations are 25cm for point D and 35cm for point B. This approach ensures an accurate assessment of the work involved in moving the charge.
shashaeee
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Calculate the work needed too move a 1.5μC test charge from point B to D.
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Can someone just check if I'm thinking correctly?

Do I just use the formula:

W = qEd ?

For the E, I calculate the magnitude and direction of the electric field at B.
And I use 25cm for the distance and q = 1.5μC
 
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W = qEd only works (pun?) if E is uniform. Instead, relate the work to electric potential or electric potential energy.
 
Ohhh, I see. So basically, the work needed to bring 1.5μC from a distance (25cm) away to a point 25cm from a charge 75μC?

Would it make sense to use this formula?


W = q(Va - Vb)
W = q[(k*Qa/ra) - (k*Qb/rb)]

q = 1.5μC
Qa = 75μC
Qb = 1.5μC

ra(75μC to D) = 25cm
rb(B to D) = 25cm
 
That's more like it. But, since the charge is being carried from B to D, you'll need to find the potential difference in going from B to D. How would you find the potential at point B due to the two 75 μC charges? Point D?
 
ohhh, righttt.

So, rewriting it ..

W = q(VD - VB)

VD = VD1 + VD2

VB = VB1 + VB2

where the distance for VD is 25cm
and the distance for VB is 35cm
 
Yes, Good!
 
oh, thank goodness lol! Thank youu!
 
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