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Let's walk through the steps to determine the net electric field at point [tex]\( P \)[/tex] given the electric fields due to charges [tex]\( A \)[/tex] and [tex]\( B \)[/tex].
1. Identify the given electric fields:
- The electric field at point [tex]\( P \)[/tex] due to charge [tex]\( A \)[/tex] is [tex]\( 8.7 \times 10^6 \)[/tex] newtons per coulomb [tex]\( (N/C) \)[/tex].
- The electric field at point [tex]\( P \)[/tex] due to charge [tex]\( B \)[/tex] is [tex]\( 5.5 \times 10^6 \)[/tex] newtons per coulomb [tex]\( (N/C) \)[/tex].
2. Add the two electric fields vectorially:
- Since the directions of the electric fields are not specified, we assume they are in the same direction (to simplify the problem). When electric fields are in the same direction, we can add their magnitudes directly.
- Calculate the sum of the magnitudes of the electric fields:
[tex]\[ E_{\text{net}} = E_A + E_B \][/tex]
3. Substitute the values:
- [tex]\( E_{\text{net}} = 8.7 \times 10^6 \, N/C + 5.5 \times 10^6 \, N/C \)[/tex]
4. Perform the addition:
- [tex]\( E_{\text{net}} = 8700000 \, N/C + 5500000 \, N/C \)[/tex]
- [tex]\( E_{\text{net}} = 14200000 \, N/C \)[/tex]
5. Convert the result back to scientific notation:
- [tex]\( E_{\text{net}} = 1.42 \times 10^7 \, N/C \)[/tex]
Therefore, the net electric field at point [tex]\( P \)[/tex] is:
D. [tex]\( 1.42 \times 10^7 \)[/tex] newtons per coulomb
1. Identify the given electric fields:
- The electric field at point [tex]\( P \)[/tex] due to charge [tex]\( A \)[/tex] is [tex]\( 8.7 \times 10^6 \)[/tex] newtons per coulomb [tex]\( (N/C) \)[/tex].
- The electric field at point [tex]\( P \)[/tex] due to charge [tex]\( B \)[/tex] is [tex]\( 5.5 \times 10^6 \)[/tex] newtons per coulomb [tex]\( (N/C) \)[/tex].
2. Add the two electric fields vectorially:
- Since the directions of the electric fields are not specified, we assume they are in the same direction (to simplify the problem). When electric fields are in the same direction, we can add their magnitudes directly.
- Calculate the sum of the magnitudes of the electric fields:
[tex]\[ E_{\text{net}} = E_A + E_B \][/tex]
3. Substitute the values:
- [tex]\( E_{\text{net}} = 8.7 \times 10^6 \, N/C + 5.5 \times 10^6 \, N/C \)[/tex]
4. Perform the addition:
- [tex]\( E_{\text{net}} = 8700000 \, N/C + 5500000 \, N/C \)[/tex]
- [tex]\( E_{\text{net}} = 14200000 \, N/C \)[/tex]
5. Convert the result back to scientific notation:
- [tex]\( E_{\text{net}} = 1.42 \times 10^7 \, N/C \)[/tex]
Therefore, the net electric field at point [tex]\( P \)[/tex] is:
D. [tex]\( 1.42 \times 10^7 \)[/tex] newtons per coulomb
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