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URGGHH!!! help

a +15 uC point charge Q1, is at a distance of 30mm from a -30uC charge Q2

1.Show that electric potential is zero at a point between the two charges which is 10mm from Q1 and 20mm and Q2. (I did this)
part 2. Calculate the electric field strength at this position and state its direction??? the answer is 2.0 x 10^9 Vm^-1 help??

calculate the electric field strength at this position and state its direction
Original post by calster102
a +15 uC point charge Q1, is at a distance of 30mm from a -30uC charge Q2

1.Show that electric potential is zero at a point between the two charges which is 10mm from Q1 and 20mm and Q2. (I did this)
part 2. Calculate the electric field strength at this position and state its direction??? the answer is 2.0 x 10^9 Vm^-1 help??

calculate the electric field strength at this position and state its direction


If you know the formula for field strength then apply it at that point for the two charges separately. Then add the two values together as vectors. Just note the direction of the fields in each case. If they are in the same direction they add. If in opposite directions they subtract. (If at an angle you can use a vector triangle method.)
You are finding the resultant as a vector.
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calster102
OP
Original post by Stonebridge
If you know the formula for field strength then apply it at that point for the two charges separately. Then add the two values together as vectors. Just note the direction of the fields in each case. If they are in the same direction they add. If in opposite directions they subtract. (If at an angle you can use a vector triangle method.)
You are finding the resultant as a vector.


you obviously get a negative field when dealing with the negative charge..i did (1.3 x 10^9)-(-6.7 x 10^8) = 2.0 x 10^9 is this exactly how you'd do it? i've subtracted surely you'd have to add them because both fields are in same direction???
(edited 9 years ago)
Original post by calster102
you obviously get a negative field when dealing with the negative charge..i did (1.3 x 10^9)-(-6.7 x 10^8) = 2.0 x 10^9 is this exactly how you'd do it? i've subtracted surely you'd have to add them because both fields are in same direction???


Yes. At some point between a positive (on the left, say) and a negative (on the right) charge, the field due to the positive charge (repulsion) points to the right and the field due to the negative charge (attraction) also points to the right.
As they both act along the same straight line you just add the magnitudes together (ignore the positive and negative from the calculation) and point out what direction that resultant is in.
Reply 4
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calster102
OP
Original post by Stonebridge
Yes. At some point between a positive (on the left, say) and a negative (on the right) charge, the field due to the positive charge (repulsion) points to the right and the field due to the negative charge (attraction) also points to the right.
As they both act along the same straight line you just add the magnitudes together (ignore the positive and negative from the calculation) and point out what direction that resultant is in.


So whenever doing these sorts of calculations just ignore any minuses?? and think about which direction the fields are..say if both were positive either side...they would be acting in opposite directions so would you subtract??
Original post by calster102
So whenever doing these sorts of calculations just ignore any minuses?? and think about which direction the fields are..say if both were positive either side...they would be acting in opposite directions so would you subtract??


Yes. I did state that in my first post.

With these questions, it's fairly straightforward if you
a. determine which direction the forces are in.
b. use the formula to calculate the field strength.
c. ignoring minus or plus in the answers to b, add the forces as vectors, taking account of the directions. Same direction add. Opposite directions subtract.
d. work out the direction of the resultant from which is the greater force.

This works if the point is on the line between the two charges.
If it's somewhere else you have to draw the forces as vectors (work out the direction first) and add using vector addition. (Vector triangle.)
Reply 6
Avatar for calster102
calster102
OP
Original post by Stonebridge
Yes. I did state that in my first post.

With these questions, it's fairly straightforward if you
a. determine which direction the forces are in.
b. use the formula to calculate the field strength.
c. ignoring minus or plus in the answers to b, add the forces as vectors, taking account of the directions. Same direction add. Opposite directions subtract.
d. work out the direction of the resultant from which is the greater force.

This works if the point is on the line between the two charges.
If it's somewhere else you have to draw the forces as vectors (work out the direction first) and add using vector addition. (Vector triangle.)

thanks

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