Can Homework Help Calculate the Electric Field from Two Point Charges?

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Electric fields are vector quantities, meaning they have both magnitude and direction. When calculating the electric field from two point charges, it's crucial to consider the signs and magnitudes of the charges involved. For example, with charges of q1=9μC and q2=-1μC, a naive calculation might incorrectly suggest that the electric field is zero. This highlights the importance of vector addition in determining the resultant electric field. Understanding these principles is essential for accurate calculations in electrostatics.
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Homework Statement
I am having trouble solving this question related to electric field.
Relevant Equations
I used k(q1/a^2 +q2/b^2) a=4.24 b=4 q1=1.6e-6 q2=2.4e-6
The answer I got is 2149 N/C (wrong)
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Hint: Electric field is a vector quantity.
 
To illustrate @TSny's point, if the charges had been ##q_1=9\mu C, q_2=-1\mu C##, your calculation would have said the field is zero.
 
Thread 'Chain falling out of a horizontal tube onto a table'

Thread 'Chain falling out of a horizontal tube onto a table'

My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...
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